first order differential equation problem solving

Module 4: Differential Equations

First-order differential equations, learning outcomes.

• Write a first-order linear differential equation in standard form
• Find an integrating factor and use it to solve a first-order linear differential equation

See the example on the introduction page for a first-order linear differential equation.

A first-order differential equation is linear if it can be written in the form

where [latex]a\left(x\right),b\left(x\right)[/latex], and [latex]c\left(x\right)[/latex] are arbitrary functions of [latex]x[/latex].

Remember that the unknown function [latex]y[/latex] depends on the variable [latex]x[/latex]; that is, [latex]x[/latex] is the independent variable and [latex]y[/latex] is the dependent variable. Some examples of first-order linear differential equations are

Examples of first-order nonlinear differential equations include

These equations are nonlinear because of terms like [latex]{\left({y}^{\prime }\right)}^{4},{y}^{3}[/latex], etc. Due to these terms, it is impossible to put these equations into the same form as the definition.

Standard Form

Consider the differential equation

Our main goal in this section is to derive a solution method for equations of this form. It is useful to have the coefficient of [latex]{y}^{\prime }[/latex] be equal to [latex]1[/latex]. To make this happen, we divide both sides by [latex]3{x}^{2}-4[/latex].

This is called the standard form of the differential equation. We will use it later when finding the solution to a general first-order linear differential equation. Returning to the definition, we can divide both sides of the equation by [latex]a\left(x\right)[/latex]. This leads to the equation

Now define [latex]p\left(x\right)=\frac{b\left(x\right)}{a\left(x\right)}[/latex] and [latex]q\left(x\right)=\frac{c\left(x\right)}{a\left(x\right)}[/latex]. Then the definition becomes

We can write any first-order linear differential equation in this form, and this is referred to as the standard form for a first-order linear differential equation.

Example: Writing First-Order Linear Equations in Standard Form

Put each of the following first-order linear differential equations into standard form. Identify [latex]p\left(x\right)[/latex] and [latex]q\left(x\right)[/latex] for each equation.

• [latex]y^{\prime} =3x - 4y[/latex]
• [latex]\frac{3xy^{\prime} }{4y - 3}=2[/latex] (here [latex]x>0[/latex])
• [latex]y=3y^{\prime} -4{x}^{2}+5[/latex]

Finally, divide both sides by [latex]3x[/latex] to make the coefficient of [latex]y^{\prime} [/latex] equal to [latex]1\text{:}[/latex]

Next divide both sides by [latex]3\text{:}[/latex]

Watch the following video to see the worked solution to Example: Writing First-Order Linear Equations in Standard Form.

Put the equation [latex]\frac{\left(x+3\right)y^{\prime} }{2x - 3y - 4}=5[/latex] into standard form and identify [latex]p\left(x\right)[/latex] and [latex]q\left(x\right)[/latex].

Multiply both sides by the common denominator, then collect all terms involving [latex]y[/latex] on one side.

[latex]y^{\prime} +\frac{15}{x+3}y=\frac{10x - 20}{x+3};p\left(x\right)=\frac{15}{x+3}[/latex] and [latex]q\left(x\right)=\frac{10x - 20}{x+3}[/latex]

Integrating Factors

We now develop a solution technique for any first-order linear differential equation. We start with the standard form of a first-order linear differential equation:

The first term on the left-hand side of [latex]{y}^{\prime }+\frac{b\left(x\right)}{a\left(x\right)}y=\frac{c\left(x\right)}{a\left(x\right)}[/latex] is the derivative of the unknown function, and the second term is the product of a known function with the unknown function. This is somewhat reminiscent of the power rule from the Differentiation Rules section. If we multiply [latex]{y}^{\prime }+p\left(x\right)y=q\left(x\right)[/latex] by a yet-to-be-determined function [latex]\mu \left(x\right)[/latex], then the equation becomes

The left-hand side of [latex]y^{\prime} +p\left(x\right)y=q\left(x\right)[/latex] can be matched perfectly to the product rule:

Matching term by term gives [latex]y=f\left(x\right),g\left(x\right)=\mu \left(x\right)[/latex], and [latex]{g}^{\prime }\left(x\right)=\mu \left(x\right)p\left(x\right)[/latex]. Taking the derivative of [latex]g\left(x\right)=\mu \left(x\right)[/latex] and setting it equal to the right-hand side of [latex]{g}^{\prime }\left(x\right)=\mu \left(x\right)p\left(x\right)[/latex] leads to

This is a first-order, separable differential equation for [latex]\mu \left(x\right)[/latex]. We know [latex]p\left(x\right)[/latex] because it appears in the differential equation we are solving. Separating variables and integrating yields

Here [latex]{C}_{2}[/latex] can be an arbitrary (positive or negative) constant. This leads to a general method for solving a first-order linear differential equation. We first multiply both sides of [latex]{y}^{\prime }+p\left(x\right)y=q\left(x\right)[/latex] by the integrating factor [latex]\mu \left(x\right)[/latex]. This gives

The left-hand side of the above equation can be rewritten as [latex]\frac{d}{dx}\left(\mu \left(x\right)y\right)[/latex].

Next integrate both sides with respect to [latex]x[/latex].

Divide both sides by [latex]\mu \left(x\right)\text{:}[/latex]

Since [latex]\mu \left(x\right)[/latex] was previously calculated, we are now finished. An important note about the integrating constant [latex]C\text{:}[/latex] It may seem that we are inconsistent in the usage of the integrating constant. However, the integral involving [latex]p\left(x\right)[/latex] is necessary in order to find an integrating factor for [latex]{y}^{\prime }+\frac{b\left(x\right)}{a\left(x\right)}y=\frac{c\left(x\right)}{a\left(x\right)}[/latex]. Only one integrating factor is needed in order to solve the equation; therefore, it is safe to assign a value for [latex]C[/latex] for this integral. We chose [latex]C=0[/latex]. When calculating the integral inside the brackets in [latex]\frac{d}{dx}\left(\mu \left(x\right)y\right)=\mu \left(x\right)q\left(x\right)[/latex], it is necessary to keep our options open for the value of the integrating constant, because our goal is to find a general family of solutions to [latex]{y}^{\prime }+\frac{b\left(x\right)}{a\left(x\right)}y=\frac{c\left(x\right)}{a\left(x\right)}[/latex]. This integrating factor guarantees just that.

Problem-Solving Strategy: Solving a First-order Linear Differential Equation

• Put the equation into standard form and identify [latex]p\left(x\right)[/latex] and [latex]q\left(x\right)[/latex].
• Calculate the integrating factor [latex]\mu \left(x\right)={e}^{\displaystyle\int p\left(x\right)dx}[/latex].
• Multiply both sides of the differential equation by [latex]\mu \left(x\right)[/latex].
• Integrate both sides of the equation obtained in step [latex]3[/latex], and divide both sides by [latex]\mu \left(x\right)[/latex].
• If there is an initial condition, determine the value of [latex]C[/latex].

Example: Solving a First-order Linear Equation

Find a general solution for the differential equation [latex]xy^{\prime} +3y=4{x}^{2}-3x[/latex]. Assume [latex]x>0[/latex].

• The integrating factor is [latex]\mu \left(x\right)={e}^{\displaystyle\int \left(\frac{3}{x}\right)dx}={e}^{3\text{ln}x}={x}^{3}[/latex].
• There is no initial value, so the problem is complete.

You may have noticed the condition that was imposed on the differential equation; namely, [latex]x>0[/latex]. For any nonzero value of [latex]C[/latex], the general solution is not defined at [latex]x=0[/latex]. Furthermore, when [latex]x<0[/latex], the integrating factor changes. The integrating factor is given by [latex]\mu \left(x\right){y}^{\prime }+\mu \left(x\right)p\left(x\right)y=\mu \left(x\right)q\left(x\right)[/latex] as [latex]f\left(x\right)={e}^{\displaystyle\int p\left(x\right)dx}[/latex]. For this [latex]p\left(x\right)[/latex] we get

since [latex]x<0[/latex]. The behavior of the general solution changes at [latex]x=0[/latex] largely due to the fact that [latex]p\left(x\right)[/latex] is not defined there.

Watch the following video to see the worked solution to Example: Solving a First-Order Linear Equation.

You can view the transcript for this segmented clip of “4.5.2” here (opens in new window) .

Find the general solution to the differential equation [latex]\left(x - 2\right)y^{\prime} +y=3{x}^{2}+2x[/latex]. Assume [latex]x>2[/latex].

Use the method outlined in the problem-solving strategy for first-order linear differential equations.

[latex]y=\frac{{x}^{3}+{x}^{2}+C}{x - 2}[/latex]

Now we use the same strategy to find the solution to an initial-value problem.

Example: A First-order Linear Initial-Value Problem

Solve the initial-value problem

• This differential equation is already in standard form with [latex]p\left(x\right)=3[/latex] and [latex]q\left(x\right)=2x - 1[/latex].
• The integrating factor is [latex]\mu \left(x\right)={e}^{\displaystyle\int 3dx}={e}^{3x}[/latex].

Integrate both sides of the equation:

Therefore the solution to the initial-value problem is

Watch the following video to see the worked solution to Example: A First-Order Linear Initial-Value Problem.

Solve the initial-value problem [latex]y^{\prime} -2y=4x+3, y\left(0\right)=-2[/latex].

[latex]y=-2x - 4+2{e}^{2x}[/latex]

• 4.5.1. Authored by : Ryan Melton. License : CC BY: Attribution
• 4.5.2. Authored by : Ryan Melton. License : CC BY: Attribution
• Calculus Volume 2. Authored by : Gilbert Strang, Edwin (Jed) Herman. Provided by : OpenStax. Located at : https://openstax.org/books/calculus-volume-2/pages/1-introduction . License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike . License Terms : Access for free at https://openstax.org/books/calculus-volume-2/pages/1-introduction

Privacy Policy

First Order Linear Differential Equations

You might like to read about Differential Equations and Separation of Variables first!

A Differential Equation is an equation with a function and one or more of its derivatives :

Here we will look at solving a special class of Differential Equations called First Order Linear Differential Equations

First Order

They are "First Order" when there is only dy dx , not d 2 y dx 2 or d 3 y dx 3 etc

A first order differential equation is linear when it can be made to look like this:

dy dx + P(x)y = Q(x)

Where P(x) and Q(x) are functions of x.

To solve it there is a special method:

• We invent two new functions of x, call them u and v , and say that y=uv .
• We then solve to find u , and then find v , and tidy up and we are done!

And we also use the derivative of y=uv (see Derivative Rules (Product Rule) ):

dy dx = u dv dx + v du dx

Here is a step-by-step method for solving them:

• 2. Factor the parts involving v
• 3. Put the v term equal to zero (this gives a differential equation in u and x which can be solved in the next step)
• 4. Solve using separation of variables to find u
• 5. Substitute u back into the equation we got at step 2
• 6. Solve that to find v
• 7. Finally, substitute u and v into y = uv to get our solution!

Let's try an example to see:

  dy dx − y x = 1

First, is this linear? Yes, as it is in the form

dy dx + P(x)y = Q(x) where P(x) = − 1 x and Q(x) = 1

So let's follow the steps:

Step 1: Substitute y = uv , and   dy dx = u dv dx + v du dx

Step 2: Factor the parts involving v

Step 3: Put the v term equal to zero

Step 4: Solve using separation of variables to find u

Step 5: Substitute u back into the equation at Step 2

Step 6: Solve this to find v

Step 7: Substitute into y = uv to find the solution to the original equation.

And it produces this nice family of curves:

What is the meaning of those curves?

They are the solution to the equation   dy dx − y x = 1

In other words:

Anywhere on any of those curves the slope minus y x equals 1

Let's check a few points on the c=0.6 curve:

Estmating off the graph (to 1 decimal place):

Why not test a few points yourself? You can plot the curve here .

Perhaps another example to help you? Maybe a little harder?

  dy dx − 3y x = x

dy dx + P(x)y = Q(x) where P(x) = − 3 x and Q(x) = x

And one more example, this time even harder :

  dy dx + 2xy= −2x 3

dy dx + P(x)y = Q(x) where P(x) = 2x and Q(x) = −2x 3

Let's see ... we can integrate by parts ... which says:

∫ RS dx = R ∫ S dx − ∫ R' ( ∫ S dx ) dx

(Side Note: we use R and S here, using u and v could be confusing as they already mean something else.)

Choosing R and S is very important, this is the best choice we found:

• R = −x 2 and
• S = 2x e x 2

So let's go:

Put in R = −x 2 and S = 2x e x 2

And also R' = −2x and ∫ S dx = e x 2

And we get this nice family of curves:

Reset password New user? Sign up

Existing user? Log in

First Order Differential Equations - Problem Solving

Already have an account? Log in here.

First order differential equations are useful because of their applications in physics, engineering, etc. They can be linear , of separable , homogenous with change of variables, or exact. Each one has a structure and a method to be solved.

Change of Variables

Exact differential equations.

A homogenous equation with change of variables needs to be in the form \[\frac{dy}{dx}=F(x,y),\] with a function \(F(x,y)\) such that \(F(tx,ty)=F(x,y) \Rightarrow F(x,y)=F\big(1,\frac{y}{x}\big)\). So we can do the change of variables \(u=\frac{y}{x} \Rightarrow \frac{dy}{dx}=\frac{du}{dx}x+u\), getting the equation (most of the time, separable) \[\frac{du}{dx}x+u=F(1,u).\]

Solve the differential equation \[\frac{dy}{dx}=\frac{y(y+x)}{x^2}.\] At first we can find out that this equation is said to be homogenous: \[\frac{dy}{dx}=\frac{y(y+x)}{x^2}=\frac{ty(ty+tx)}{t^2x^2}=\frac{t^2y(y+x)}{t^2x^2}=\frac{y(y+x)}{x^2}.\] Then, we can do the change of variables: \[u=\frac{y}{x} \Rightarrow \frac{dy}{dx}=\frac{du}{dx}x+u,\] and get the following separable equation: \[\begin{align} \frac{du}{dx}x+u&=u^2+u \\\\ \frac{du}{dx}&=\frac{u^2}{x} \\\\ \int \frac{du}{u^2}&=\displaystyle \int \frac{dx}{x}\\\\ \frac{1}{u}&=-\ln(x)+C \\\\ u&=-\frac{1}{\ln(x)+C} \\\\ y&=-\frac{x}{\ln(x)+C}, \end{align}\] where \(C\) is the constant of integration. \(_\square\)

Let's take the following differential equation: \[\begin{align} A(x,y)dy+B(x,y)dx&=0\\ \Rightarrow \frac{dy}{dx}&=-\frac{B(x,y)}{A(x,y)}. \end{align}\] It would be a very nice coincidence that there was a function \(F(x,y)\) such that \[\frac{dy}{dx}=-\frac{\frac{\partial F(x,y)}{\partial x}}{\frac{\partial F(x,y)}{\partial y}}=-\frac{\partial y}{\partial x}.\] A differential equation is said to be exact iif there exists a \(C^2\) function \(F(x,y)\) such that \(A(x,y)=\frac{\partial F(x,y)}{\partial y}\) and \(B(x,y)=\frac{\partial F(x,y)}{\partial x}\). \[\] To prove that such a function exists we can use partial derivatives : \[\begin{align} A(x,y)=\frac{\partial F(x,y)}{\partial y} &\Rightarrow \frac{\partial A(x,y)}{\partial x}=\frac{\partial ^2 F(x,y)}{\partial x \partial y}\\ B(x,y)=\frac{\partial F(x,y)}{\partial x} &\Rightarrow \frac{\partial B(x,y)}{\partial y}=\frac{\partial ^2 F(x,y)}{\partial y \partial x}. \end{align}\] As \(F\) is \(C^2,\) we can conclude \[\begin{align} \frac{\partial ^2 F(x,y)}{\partial y \partial x}&=\frac{\partial ^2 F(x,y)}{\partial x \partial y}\\ \Rightarrow \frac{\partial A(x,y)}{\partial x}&=\frac{\partial B(x,y)}{\partial y}. \end{align}\] And that is the most important requirement for saying a differential equation is exact. \[\] When the function \(F\) is found, the solution to the differential equation is given, implicitly, by \[F(x,y)=0.\]

Solve the differential equation \[\ln\big(x^2y\big)dy+\frac{2\ln(y)}{x}dx=0.\] On the way to prove that is an exact equation is to take \[\begin{align} \frac{\partial}{\partial x}\big(\ln(x^2y)\big)&=\frac{\partial}{\partial y}\left(\frac{2\ln(y)}{x}\right)\\ \frac{2}{xy}&=\frac{2}{xy}. \end{align}\] So it is exact, and we know \[\begin{align} \ln(x^2y)=\frac{\partial F(x,y)}{\partial y} \Rightarrow F(x,y) &= \int \ln(x^2y)dy \\ &=y\ln(x^2y)-\displaystyle \int dy\\ &=y\big(\ln(x^2y)+1\big)+C(x) \end{align}\] and \[\begin{align} \frac{2\ln(y)}{x}&=\frac{\partial F(x,y)}{\partial x} \\ \frac{2\ln(y)}{x}&=\frac{2}{x}+C'(x)\\ C(x)&=2\ln(x)\ln(y)-2\ln(x)+C\\ F(x,y)&=y\big(\ln(x^2y)+1\big)+2\ln(x)\big(\ln(y)+1\big)+C. \end{align}\] So the solution to the differential equation is given by \[y\big(\ln(x^2y)+1\big)+2\ln(x)\big(\ln(y)+1\big)+C=0.\ _\square\]

Problem Loading...

Note Loading...

Set Loading...

First Order Differential Equation

A first order differential equation is a differential equation where the maximum order of a derivative is one and no other higher-order derivative can appear in this equation. A first-order differential equation is generally of the form F(x, y, y') = 0, where y is a dependent variable and x is an independent variable and y' appears explicitly in the differential equation. It can also be written as F(t, f(t), f'(t)) = 0, where f(t) is the solution of the differential equation. A linear first order differential equation is a differential equation with a derivative of order one and the degree of the equation is also one.

In this article, we will explore the concept of first order differential equations, ways to find their solutions, first-order initial value problem differential equations, and their applications. We will solve a few examples for a better understanding of the concept.

What is First Order Differential Equation?

A First Order Differential Equation is generally written as F(x, y, y') = 0, where y' is the first-order derivative and appears explicitly in the equation, x is an independent variable and y is a function of x. We say that the function f(t) is a solution of the first order differential equation F(t, f(t), f'(t)) = 0, for all values of t. A real-life example of the first-order differential equation is Newton's law of cooling equation given by, y' = k(M - y) and it can be expressed as F(t, y, y') = k(M - y) - y'. Let us see some other examples of the differential equations of first order:

• y' = t 2 + 1 ⇒ F(t, y, y') = t 2 + 1 - y'
• y' = 2(25 - y) ⇒ F(t, y, y') = 2(25 - y) - y'
• mv'(t) = -mg ⇒ F(t, v, v') = -mg - mv'(t)

Linear First Order Differential Equation

A linear first order differential equation is of the form y' + y P(x) = Q(x) or dy/dx + y P(x) = Q(x), where y, P, Q are functions of x, and y' is the first-order derivative of y. Such differential equations have a degree of the derivative equal to one, that is why it is called the linear first order differential equation. We can solve such equations using the method of integrating factors.

Solution First Order Differential Equation

Now, we can solve first order differential equations using different methods such as separating the variables, integrating factors method, variation of parameters, etc. We can determine a particular solution p(x) and a general solution g(x) corresponding to the homogeneous first-order differential equation y' + y P(x) = 0 and then the general solution to the non-homogeneous first order differential equation y' + y P(x) = Q(x) is given by y(x) = p(x) + g(x). Let us solve a few examples using different methods to understand the application of each method.

Separable First Order Differential Equation

The general form of the separable first order differential equation is dy/dx = f(y).g(x). Here we can separate the variables on the two sides of the equation, i.e., dy/dx = f(y).g(x) can also be written as dy/f(y) = g(x) dx by separating the variables and then we can solve the equation by integration. Let us consider an example of a first order differential equation and find its solution by the separation of variables method.

Example: Consider differential equation dy/dx = (5y + 4)x.

Keeping the variable y on the LHS and x on the RHS of the equation, we have

dy/dx = (5y + 4)x

⇒ dy/(5y + 4) = xdx

Now, integrating both sides of the equation, we have

∫dy/(5y + 4) = ∫ x dx

⇒ (1/5) ln |5y + 4| = x 2 /2 + C

⇒ ln |5y + 4| = 5 (x 2 /2 + C)

⇒ ln |5y + 4| = 5x 2 /2 + 5C

⇒ 5y + 4 = e 5x 2 /2 + 5C

⇒ y = (1/5) [e 5x 2 /2 + 5C - 4]

⇒ y = (1/5)e 5x 2 /2 e 5C - 4/5

⇒ y = Ke 5x 2 /2 - 4/5, where K = (1/5)e 5C

Hence, y = Ke 5x 2 /2 - 4/5 is the general solution of the first order differential equation dy/dx = (5y + 4)x.

Solving First Order Differential Equation Using Integrating Factors

A first order differential equation of the form dy/dx + y P(x) = Q(x) can be solved using the integrating factors method. We can follow the given steps to find the general solution of the differential equation:

• Step 1: Simplify the first order differential equation and express it as dy/dx + y P(x) = Q(x)
• Step 2: Determine the integrating factor given by, I.F. = e ∫P(x) dx
• Step 3: Multiply the differential equation dy/dx + y P(x) = Q(x) by the I.F. to obtain d(y × I.F.)/dx = Q(x) × I.F.
• Step 4: Now, integrate both sides of the equation d(y × I.F.)/dx = Q(x) × I.F. to get the general solution.

Let us solve a first order differential equation using the integrating factor method to understand its application.

Example: Consider the differential equation xy' + 3y = 4x 2 - 3x.

First, we will write the given first-order differential equation in the form y' + y P(x) = Q(x)

Dividing xy' + 3y = 4x 2 - 3x by x, we have

y' + (3/x) y = 4x - 3

Here P(x) = 3/x and Q(x) = 4x - 3

Now, the integrating factor I.F. = e ∫P(x) dx = e ∫(3/x) dx = e 3ln x = x 3

Multiplying the first-order differential equation y' + (3/x) y = 4x - 3 by the I.F. = x 3 , we have

x 3 (y' + (3/x) y) = x 3 (4x - 3)

⇒ x 3 y' + 3x 2 y = 4x 4 - 3x 3

⇒ d(yx 3 )/dx = 4x 4 - 3x 3

Integrating both sides of the equation with respect to x, we get

⇒ ∫[d(yx 3 )/dx] dx = ∫(4x 4 - 3x 3 ) dx

⇒ yx 3 = (4/5)x 5 - (3/4)x 4 + C

⇒ y = (4/5)x 2 - (3/4)x + Cx -3

Hence, the general solution of the first order differential equation xy' + 3y = 4x 2 - 3x using the integrating factor method is y = (4/5)x 2 - (3/4)x + Cx -3

Initial Value Problem First Order Differential Equation

First order initial value problem differential equation is of the form F(x, y, y') = 0 and initial value y(x 0 ) = y 0 , where x 0 is a fixed value of x and y 0 is the corresponding value of y and (x 0 ,y 0 ) satisfies the general solution y(x). The initial value of the differential equation helps to find the particular solution of the first-order differential equation. Let us consider an example to see how to determine the solution of a differential equation using the initial value.

Example: Consider the first-order differential equation y' = x 2 + 1, y(1) = 4

First, we will evaluate the general solution of the given differential equation. We have dy/dx = x 2 + 1 which can be solved by separating the variables.

dy/dx = x 2 + 1

⇒ dy = (x 2 + 1) dx

Integrating both sides of the equation, we get

∫ dy = ∫ (x 2 + 1) dx

⇒ y = x 3 /3 + x + C, which is the general solution of the given first order differential equation.

We have y(1) = 4. Substitute this in the general solution to determine the value of C, and hence the particular solution.

4 = 1 3 /3 + 1 + C

⇒ C = 4 - 1/3 - 1

Therefore y = y = x 3 /3 + x + 8/3 is the particular solution of the initial value problem differential equation y' = x 2 + 1, y(1) = 4.

Important Notes on First Order Differential Equation

• Some of the important applications of the first order differential equation are in Newton's law of cooling, growth and decay models, and electric circuits.
• First-order differential equations can be solved using the separation of variables, integrating factor, and variation of parameters method.

Related Topics on First Order Differential Equation

• Rules of Differentiation
• Product Rule Formula
• Chain Rule Formula

First Order Differential Equation Examples

Example 1: Solve the first-order differential equation x 2 y' = x 3 + 2

Solution: To solve the given differential equation, we will separate its variables.

x 2 y' = x 3 + 2

⇒ y' = x + 2/x 2

Integrating both sides w.r.t. x

⇒ ∫(dy/dx) dx = ∫ (x + 2/x 2 ) dx

⇒ y = x 2 /2 - 2/x + C

Answer: Hence the general solution is y = x 2 /2 - 2/x + C

Example 2: Determine the solution of the differential equation xy(dy/dx) + y 2 + 2y = 0 using the integrating factor method.

Solution: The given first order differential equation can be written as:

xy(dy/dx) + y 2 + 2y = 0

⇒ dy/dx + y/x + 2/x = 0 [Dividing the differential equation by xy]

⇒ dy/dx + y/x = - 2/x

On comparing the differential equation dy/dx + y/x = - 2/x with dy/dx + y P(x) = Q(x), we have P(x) = 1/x, and Q(x) = -2/x. Next, find the integrating factor.

I.F. = e ∫P(x) dx = e ∫(1/x) dx = e ln x = x

Multiplying the first-order differential equation dy/dx + y/x = - 2/x by the I.F. = x, we have

x(dy/dx + y/x) = x(-2/x)

⇒ xdy/dx + y = -2

⇒ d(xy)/dx = -2

⇒ ∫ [d(xy)/dx]dx = ∫-2 dx

⇒ xy = -2x + C

⇒ y = -2 + C/x

Answer: The general solution of xy(dy/dx) + y 2 + 2y = 0 is y = -2 + C/x.

go to slide go to slide

Book a Free Trial Class

First Order Differential Equation Questions

Faqs on first order differential equation, what is first order differential equation in calculus.

A first order differential equation is a differential equation where the maximum order of a derivative is one and no other higher-order derivative can appear in this equation. A first-order differential equation is generally of the form F(x, y, y') = 0.

What is a Linear First Order Differential Equation?

A linear first order differential equation is of the form y' + y P(x) = Q(x) or dy/dx + y P(x) = Q(x), where y, P, Q are functions of x, and y' is the first-order derivative of y. Such differential equations have a degree of the derivative equal to one, that is why it is called the linear first order differential equation.

What is a Homogeneous First Order Differential Equation?

A first order differential equation M(x, y) dx + N(x, y) dy = 0 is said to be homogeneous if both M(x, y) and N(x, y) are homogeneous. We can write a homogeneous linear first-order differential equation is of the form y' + p(x)y = 0.

How to Find the General Solution of First Order Differential Equation?

The general solution of the first order differential equation can be evaluated using the separation of variables method or using the integrating factor method.

How to Find the Integrating Factor of First Order Differential Equation?

The integrating factor of the first order differential equation dy/dx + y P(x) = Q(x) can be calculated using the formula I.F. = e ∫P(x) dx .

What is First Order First Degree Differential Equation?

First Order First Degree Differential Equation is a differential equation involving the first-order derivative and no other higher-order derivative can appear in this equation and the degree of the derivative is one.

How Do You Solve First Order Differential Equation?

The first order differential equation can be solved using various methods such as separation of variables, integrating factor or variation of parameters.

Browse Course Material

Course info, instructors.

• Prof. Arthur Mattuck
• Prof. Haynes Miller
• Dr. Jeremy Orloff
• Dr. John Lewis

Departments

• Mathematics

As Taught In

• Differential Equations
• Linear Algebra

Learning Resource Types

Problem sets.

You are leaving MIT OpenCourseWare

• Math Article

First Order Differential Equation

A first-order differential equation is defined by an equation:  dy/dx =f (x,y)  of two variables x and y with its function f(x,y) defined on a region in the xy-plane. It has only the first derivative dy/dx   so that the equation is of the first order and no higher-order derivatives exist. The differential equation in first-order can also be written as;

y’ = f (x,y)   or

(d/dx) y = f (x,y)

The differential equation is generally used to express a relation between the function and its derivatives. In Physics and chemistry, it is used as a technique for determining the functions over its domain if we know the functions and some of the derivatives.

Table of Contents:

• First Order Linear Equation
• Applications
• Solved Examples

First Order Linear Differential Equation

If the function f is a linear expression in y, then the first-order differential equation  y’ = f (x, y) is a linear equation. That is, the equation is linear and the function f takes the form

f(x,y) = p(x)y + q(x)

Since the linear equation is y = mx+b

where p and q are continuous functions on some interval I. Differential equations that are not linear are called nonlinear equations.

Consider the first-order differential equation y’ = f (x,y),  is a linear equation and it can be written in the form

• y’ + a(x)y = f(x)

where a(x) and f(x) are continuous functions of x

The alternate method to represent the first-order linear equation in a reduced form is

(dy/dx) + P(x)y = Q (x)

Where P(x) and Q(x) are the functions of x which are the continuous functions. If P(x) or Q(x) is equal to zero, the differential equation is reduced to the variable separable form. It is easy to solve when the differential equations are in variable separable form.

Types of First Order Differential Equations

There are basically five types of differential equations in the first order. They are:

• Linear Differential Equations
• Homogeneous Equations
• Exact Equations
• Separable Equations

Integrating Factor

First Order Differential Equations Solutions

Usually, there are two methods considered to solve the linear differential equation of first order.

• Using Integrating Factor
• Method of variation of constant

Let us discuss each method one by one to get the solutions for differential equations of the first order.

If a linear differential equation is written in the standard form:

y’ + a(x)y = 0

Then, the integrating factor is defined by the formula

u(x) = exp (∫a(x)dx)

Multiplying the integrating factor u(x) on the left side of the equation that converts the left side into the derivative of the product y(x)u(x).

The general solution of the differential equation is expressed as follows:

where C is an arbitrary constant.

Method of Variation of a Constant

This method is similar to the integrating factor method. Finding the general solution of the homogeneous equation is the first necessary step.

The general solution of the homogeneous equation always contains a constant of integration C. We can replace the constant C with a certain unknown function C(x). When substituting this solution into the non-homogeneous differential equation, we can be able to determine the function C(x). This approach of the algorithm is called the method of variation of a constant. However, both methods lead to the same solution.

Properties of First-order Differential Equations

The Linear first-order differential equation possesses the following properties.

• It does not have any transcendental functions like trigonometric functions and logarithmic functions.
• The products of y and any of its derivatives are not present.

Applications of First-order Differential Equation

Some of the applications which use the first-order differential equation are as follows:

• Newton’s law of cooling
• Growth and decay
• Orthogonal trajectories
• Electrical circuits
• Falling Body Problems
• Dilution Problems

Problems and Solutions

Question 1 : Solve the equation y′−y−xe x   = 0

Solution :  Given, y′−y−xe x  = 0

Rewrite the given equation and the equation becomes,

y′−y  =  xe x

Using the integrating factor, it becomes;

Therefore, the general solution of the linear equation is

Question 2: Solve the differential equation y’+2xy = x.

Solution: The given equation is already in a standard form, y’ + P(x)y = Q(x)

Therefore, P(x) = 2x and Q(x) = x

Register with BYJU’S – The learning app to get more information about the maths-related articles and start practice with the problems.

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

• School Guide
• Mathematics
• Number System and Arithmetic
• Trigonometry
• Probability
• Mensuration
• Maths Formulas
• Class 8 Maths Notes
• Class 9 Maths Notes
• Class 10 Maths Notes
• Class 11 Maths Notes
• Class 12 Maths Notes

First Order Differential Equation

A first-order differential equation is a type of differential equation that involves derivatives of the first degree (first derivatives) of a function. It does not involve higher derivatives. It can generally be expressed in the form: dy/dx = f(x, y). Here, y is a function of x, and f(x, y) is a function that involves x and y.

Table of Content

First-Order Differential Equation

Example of first-order differential equation, types of first order differential equation, first-order differential equation solution, properties of first-order differential equation, first-order differential equation formulas, applications of first-order differential equation, first order differential equation examples with solution, first order differential equation questions.

It is defined by an equation dy/dx = f (x, y) where x and y are two variables and f(x, y) are two functions. It is defined as a region in the xy plane. These types of equations have only the first derivative dy/dx, so the equation is of the first order, and no higher-order derivatives exist.

Differential equations of first order are written as;

y’ = f (x, y) (d/dx)y = f (x, y)

Let’s learn more about First-order Differential Equations, types, and examples of First-order Differential equations in detail below.

The first-order differential equation is defined by an equation: dy/dx = f(x, y). It involves two variables x and y, where the function f(x, y) is defined on a region in the xy-plane. For any linear expression in y, the first-order differential equation [Tex]y’ = f (x, y)[/Tex] is linear. Nonlinear differential equations are those that aren’t linear.

Check: Differential Equations | Definition, Formula, Types, Examples

Some examples of first-order differential equation

• dy/dx = x + 11
• dy/dx = 4x – 5

This equation represents a first-order ordinary differential equation where the derivative of y concerning x is equal to 2x.

First-order differential Equations are classified into several forms, each having its characteristics. Types of the First-Order Differential Equations:

• Linear Differential Equations
• Homogeneous Equations
• Exact Equations
• Separable Equations
• Non-Linear First Order Differential Equations

Linear Differential Equation

A linear differential equation consists of a variable, its derivative, and additional functions. It’s expressed in the standard form as:

dy/dx + P(x)y = Q(x)

• P(x) and Q(x) may be functions of x or numerical constants.

Homogenous First Order Differential Equation

A  homogeneous differential equation  is a function of the form (f(x,y) \frac{dy}{dx} = g(x,y)), where the degree of (f) and (g) is the same. A function (F(x,y)) can be considered homogeneous if it satisfies the condition: (F(\lambda x, \lambda y) = \lambda^n F(x,y)) for any nonzero constant (\lambda).

• In simpler terms, a differential equation is homogeneous if it involves a function and its derivatives in a way that maintains a consistent degree.

Example of Homogenous First Order Differential Equation

Consider the differential equation: (\frac{dy}{dx} = \frac{x^2 – y^2}{xy}). This equation is homogeneous because both the numerator and denominator have the same degree (1).

Exact Differential Equations

The formula Q (x,y) dy + P (x,y) dx = 0 is considered to be an exact differential equation if a function f of two variables, x and y, exists that has continuous partial derivatives and can be divided into the following categories.

The general solution of the equation is:

u(x, y) = C

Since, ux(x, y) = p(x, y) and uy (x, y) = Q(x, y)

where, C is Constant of Integration

Separable Differential Equations

Separable differential equations are a special type of differential equations where the variables involved can be separated to find the solution of the equation. Separable differential equations can be written in the form of:

dy/dx = f(x) × g(y) where x and y are the variables and are explicitly separated from each other.

Once the variables have been separated, it is simple to find the differential equation’s solution by integrating both sides of the equation. After the variables are separated, the separable differential equation

dy/dx = f(x) × g(y) is expressed as dy/g(y) = f(x) × dx

Non Linear First Order Differential Equation

Nonlinear first-order differential equations do not fit the linear form and can involve powers or products of y and its derivatives. For example:

dy/dx = y 2 – x

First-Order Differential Equation is generally solved and simplified using two methods mentioned below.

• Using Integrating Factor
• Method of variation of constant

Integrating Factor Homogenous Differential Equation

The  integrating factor  is a function used to solve first-order differential equations. It is most commonly applied to  ordinary linear differential equations of the first order . If a linear differential equation is written in the standard form y’ + a(x)y = 0

Then, the integrating factor (μ) is defined as: [Tex](\mu = e^{\int P(x)dx})[/Tex]

Solution using Integration Factor

Using integrating factor can be used to simplify and facilitate the solution of linear differential equations. The entire equation becomes exact when the integrating factor, which is a function of x, is multiplied.

For any linear differential equation is written in the standard form as:

y’ + a(x)y = 0

Then, the integrating factor is defined as:

u(x) = e (∫a(x)dx)

Multiplication of integrating factor u(x) to the left side of the equation converts the left side into the derivative of the product y(x).u(x). General solution of the differential equation is:

y = {∫u(x).f(x)dx + c}/u(x)

where C is an arbitrary constant.

Method of Variation of a Constant

Method of Variation of a Constant is a similar method to solve first order differential equation. In first step, we need to do y’ + a(x)y = 0 . In this method of solving first order differential equation, homogeneous equation always contains a constant of integration C.

• In solving non-homogeneous linear differential equations, we replace the constant (C) with an unknown function (C(x)).
• By substituting this solution into the non-homogeneous equation, we can determine the specific form of the function (C(x)).
• This approach allows us to find a particular solution that satisfies the given non-homogeneous equation.
• Interestingly, both the method of variation of constants and the method of integrating factors lead to the same solution.

Remember, this technique helps us handle non-homogeneous differential equations by introducing a function that varies with the independent variable!

Various properties of linear first-order differential equation are:

• No transcendental functions like trigonometric functions and logarithmic functions are used in linear first-order differential equation.
• In linear first-order differential equation, products of y and any of its derivatives are not present.

[Tex]\begin{array}{|c|c|c|} \hline \textbf{Type of Equation} & \textbf{General Form} & \textbf{Solution Method} \\ \hline \text{Linear Differential Equations} & \frac{dy}{dx} + p(x)y = q(x) & \text{Use an integrating factor, } \mu(x) = e^{\int p(x) \, dx}, \text{ then solve } \frac{d}{dx}(\mu(x)y) = \mu(x)q(x). \\ \hline \text{Homogeneous Equations} & \frac{dy}{dx} = f\left(\frac{y}{x}\right) & \text{Make the substitution } v = \frac{y}{x}, \text{ then solve the resulting separable differential equation for } v. \\ \hline \text{Exact Equations} & M(x,y) + N(x,y)\frac{dy}{dx} = 0 & \text{Find a potential function } \Psi \text{ such that } d\Psi = Mdx + Ndy, \text{ where } \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. \\ \hline \text{Separable Equations} & \frac{dy}{dx} = g(x)h(y) & \text{Separate variables and integrate: } \int \frac{1}{h(y)} dy = \int g(x) dx + C. \\ \hline \text{Integrating Factor} & \text{Often used for non-exact linear equations} & \text{Find an integrating factor } \mu(x) \text{ that makes the equation exact, then proceed as with exact equations.} \\ \hline \end{array} [/Tex]

Numerous disciplines, including physics, engineering, biology, economics, and more, first-order differential equations are used. Among other things, they are used to simulate phenomena like fluid dynamics, electrical circuits, population dynamics, and chemical reactions. Various applications of the first-order differential equation are:

• Used in Newton’s Law of Cooling
• Used in Orthogonal Trajectories
• Used in Falling Body Problems
• Used in Dilution Problems
• Calculus in Maths
• How to solve Differential Equations?
• Application of Differential Equation

Below are the example of problems on First Order Differential Equation.

Example 1: Solve the following separable differential equation: dy/dx = x/y 2

First, we separate the variables: y 2. dy = x.dx Integrating both sides: ∫y 2. dy = ∫x.dx y 3 /3 = x 2 /2 + C

Example 2: Solve the following linear differential equation: dy/dx + 2xy – x = 0

Equation in the standard form: dy/dx + 2xy – x = 0 Now, we can use an integrating factor to solve it: f(x) = e ∫2xdx f(x) = [Tex]e^{x^2}[/Tex] Multiplying both sides by the integrating factor: [Tex]e^{x^2} \frac{dy}{dx} + 2xye^{x^2} – xe^{x^2} = 0[/Tex] [Tex]\frac{d}{dx} (ye^{x^2}) – xe^{x^2} = 0[/Tex] Integrating both sides: [Tex]ye^{x^2} = \frac{x^2}{2} + C[/Tex]

Example 3: Solve the first-order differential equation x 3 y’ = x + 2

x 3 y’ = x + 2 ⇒ y’ = (x + 2)/x 3 Integrating both sides w.r.t. x ⇒ ∫(dy/dx) dx = ∫ {(x + 2)/x 3 } dx ⇒ y = -1/x – 1/x 2 + C

1. For differential equation dy/dx + yx 2 = sin x find integrating factor.

2. Find the solution of differential equation dy/dx = y 2 (x 2 +1).

3. Solve the differential equation dy/dx + 2x 3 y = x.

First-order differential equations involve the first derivative of an unknown function and are fundamental in modeling various dynamic systems. These equations can typically be expressed in the form [Tex]\frac{dy}{dx} = f(x, y)[/Tex] . There are several methods to solve first-order differential equations, including separation of variables, integrating factors, and exact equations. In separation of variables, the equation is rearranged so that all terms involving x are on one side and all terms involving y are on the other, allowing integration of both sides.

First Order Differential Equation – FAQs

What is first order differential equation.

A first order differential equation is a differential equation where the maximum order of a derivative is one and no other higher-order derivative can appear in this equation. A first-order differential equation is generally of the form [Tex]dy/dx =f (x,y)[/Tex] .

What are the types of First Order Differential Equations?

First Order Differential Equations are classified into three categories: (i) Separable Differential Equation, (ii) Linear Differential Equation and (iii) Exact Differential Equation

Give one example of First Order Differential Equation.

An Example of First Order Differential Equations: (dy/dx = 2x)

What are application of First Order Differential Equations?

First Order Differential Equation are used in various fields like physics, engineering, biology, applied mathematics, etc.

What is Homogeneous First Order Differential Equation?

A first order differential equation M(x, y) dx + N(x, y) dy = 0 is said to be homogeneous if both M(x, y) and N(x, y) are homogeneous. We can write a homogeneous linear first-order differential equation is of the form y’ + p(x)y = 0.

Please Login to comment...

Similar reads.

• School Learning
• Maths-Class-12
• Discord Emojis List 2024: Copy and Paste
• Best Adblockers for Twitch TV: Enjoy Ad-Free Streaming in 2024
• PS4 vs. PS5: Which PlayStation Should You Buy in 2024?
• Best Mobile Game Controllers in 2024: Top Picks for iPhone and Android
• System Design Netflix | A Complete Architecture

Improve your Coding Skills with Practice

What kind of Experience do you want to share?

• Solutions Integral Calculator Derivative Calculator Algebra Calculator Matrix Calculator More...
• Graphing Line Graph Exponential Graph Quadratic Graph Sine Graph More...
• Calculators BMI Calculator Compound Interest Calculator Percentage Calculator Acceleration Calculator More...
• Geometry Pythagorean Theorem Calculator Circle Area Calculator Isosceles Triangle Calculator Triangles Calculator More...
• Tools Notebook Groups Cheat Sheets Worksheets Study Guides Practice Verify Solution

• Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Number Line Expanded Form Mean, Median & Mode
• Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi (Product) Notation Induction Prove That Logical Sets Word Problems
• Pre Calculus Equations Inequalities Scientific Calculator Scientific Notation Arithmetics Complex Numbers Coterminal Angle Polar/Cartesian Simultaneous Equations System of Inequalities Polynomials Rationales Functions Arithmetic & Comp. Coordinate Geometry Plane Geometry Solid Geometry Conic Sections Trigonometry
• Calculus Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Fourier Transform
• Functions Line Equations Functions Arithmetic & Comp. Conic Sections Transformation
• Linear Algebra Matrices Vectors
• Trigonometry Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify
• Statistics Mean Geometric Mean Quadratic Mean Average Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge Standard Normal Distribution
• Physics Mechanics
• Chemistry Chemical Reactions Chemical Properties
• Finance Simple Interest Compound Interest Present Value Future Value
• Economics Point of Diminishing Return
• Conversions Roman Numerals Radical to Exponent Exponent to Radical To Fraction To Decimal To Mixed Number To Improper Fraction Radians to Degrees Degrees to Radians Hexadecimal Scientific Notation Distance Weight Time Volume
• Pre Algebra
• Pre Calculus
• First Derivative
• Product Rule
• Quotient Rule
• Sum/Diff Rule
• Second Derivative
• Third Derivative
• Higher Order Derivatives
• Derivative at a point
• Partial Derivative
• Implicit Derivative
• Second Implicit Derivative
• Derivative using Definition
• Slope of Tangent
• Curved Line Slope
• Extreme Points
• Tangent to Conic
• Linear Approximation
• Difference Quotient
• Horizontal Tangent
• One Variable
• Multi Variable Limit
• At Infinity
• L'Hopital's Rule
• Squeeze Theorem
• Substitution
• Sandwich Theorem
• Indefinite Integrals
• Definite Integrals
• Partial Fractions
• U-Substitution
• Trigonometric Substitution
• Weierstrass Substitution
• Long Division
• Improper Integrals
• Antiderivatives
• Double Integrals
• Triple Integrals
• Multiple Integrals
• Limit of Sum
• Area under curve
• Area between curves
• Area under polar curve
• Volume of solid of revolution
• Function Average
• Riemann Sum
• Trapezoidal
• Simpson's Rule
• Midpoint Rule
• Geometric Series Test
• Telescoping Series Test
• Alternating Series Test
• P Series Test
• Divergence Test
• Comparison Test
• Limit Comparison Test
• Integral Test
• Absolute Convergence
• Radius of Convergence
• Interval of Convergence
• Linear First Order
• Linear w/constant coefficients
• Second Order
• Non Homogenous
• System of ODEs
• IVP using Laplace
• Series Solutions
• Method of Frobenius
• Gamma Function
• Taylor Series
• Maclaurin Series
• Fourier Series
• Fourier Transform
• Linear Algebra
• Trigonometry
• Conversions

Number Line

• 2y'-y=4\sin(3t)
• ty'+2y=t^2-t+1
• y'=e^{-y}(2x-4)
• \frac{dr}{d\theta}=\frac{r^2}{\theta}
• y'+\frac{4}{x}y=x^3y^2
• y'+\frac{4}{x}y=x^3y^2, y(2)=-1
• laplace\:y^{\prime}+2y=12\sin(2t),y(0)=5
• bernoulli\:\frac{dr}{dθ}=\frac{r^2}{θ}
• How do you calculate ordinary differential equations?
• To solve ordinary differential equations (ODEs), use methods such as separation of variables, linear equations, exact equations, homogeneous equations, or numerical methods.
• Which methods are used to solve ordinary differential equations?
• There are several methods that can be used to solve ordinary differential equations (ODEs) to include analytical methods, numerical methods, the Laplace transform method, series solutions, and qualitative methods.
• Is there an app to solve differential equations?
• To solve ordinary differential equations (ODEs) use the Symbolab calculator. It can solve ordinary linear first order differential equations, linear differential equations with constant coefficients, separable differential equations, Bernoulli differential equations, exact differential equations, second order differential equations, homogenous and non homogenous ODEs equations, system of ODEs, ODE IVP's with Laplace Transforms, series solutions to differential equations, and more.
• What is the difference between ODE and PDE?
• An ordinary differential equation (ODE) is a mathematical equation involving a single independent variable and one or more derivatives, while a partial differential equation (PDE) involves multiple independent variables and partial derivatives. ODEs describe the evolution of a system over time, while PDEs describe the evolution of a system over space and time.
• What is the importance of studying ordinary differential equations?
• Ordinary differential equations (ODEs) help us understand and predict the behavior of complex systems, and for that, it is a fundamental tool in mathematics and physics.
• What is the numerical method?
• Numerical methods are techniques for solving problems using numerical approximation and computation. These methods are used to find approximate solutions to problems that cannot be solved exactly, or for which an exact solution would be difficult to obtain.

ordinary-differential-equation-calculator

• Advanced Math Solutions – Ordinary Differential Equations Calculator <div class="p1"> Differential equations contain derivatives, solving the equation involves integration (to get...

We want your feedback

Please add a message.

Message received. Thanks for the feedback.

• Dean’s Office
• External Advisory Council
• Computing Council
• Extended Computing Council
• Undergraduate Advisory Group
• Break Through Tech AI
• Building 45 Event Space
• Infinite Mile Awards: Past Winners
• Frequently Asked Questions
• Undergraduate Programs
• Graduate Programs
• Educating Computing Bilinguals
• Online Learning
• Industry Programs
• AI Policy Briefs
• Envisioning the Future of Computing Prize 2024
• SERC Symposium 2023
• SERC Case Studies
• SERC Scholars Program
• SERC Group Leaders
• Common Ground Subjects
• For First-Year Students and Advisors
• For Instructors: About Common Ground Subjects
• Common Ground Award for Excellence in Teaching
• New & Incoming Faculty
• Faculty Resources
• Faculty Openings
• Search for: Search
• MIT Homepage

A framework for solving parabolic partial differential equations

A new algorithm solves complicated partial differential equations by breaking them down into simpler problems, potentially guiding computer graphics and geometry processing.


Computer graphics and geometry processing research provide the tools needed to simulate physical phenomena like fire and flames, aiding the creation of visual effects in video games and movies as well as the fabrication of complex geometric shapes using tools like 3D printing.

Under the hood, mathematical problems called partial differential equations (PDEs) model these natural processes. Among the many PDEs used in physics and computer graphics, a class called second-order parabolic PDEs explain how phenomena can become smooth over time. The most famous example in this class is the heat equation, which predicts how heat diffuses along a surface or in a volume over time.

Researchers in geometry processing have designed numerous algorithms to solve these problems on curved surfaces, but their methods often apply only to linear problems or to a single PDE. A more general approach by researchers from MIT’s Computer Science and Artificial Intelligence Laboratory (CSAIL) tackles a general class of these potentially nonlinear problems.

In a paper recently published in the Transactions on Graphics journal and presented at the SIGGRAPH conference, they describe an algorithm that solves different nonlinear parabolic PDEs on triangle meshes by splitting them into three simpler equations that can be solved with techniques graphics researchers already have in their software toolkit. This framework can help better analyze shapes and model complex dynamical processes.

“We provide a recipe: If you want to numerically solve a second-order parabolic PDE, you can follow a set of three steps,” says lead author Leticia Mattos Da Silva SM ’23, an MIT PhD student in electrical engineering and computer science (EECS) and CSAIL affiliate. “For each of the steps in this approach, you’re solving a simpler problem using simpler tools from geometry processing, but at the end, you get a solution to the more challenging second-order parabolic PDE.”

To accomplish this, Da Silva and her coauthors used Strang splitting, a technique that allows geometry processing researchers to break the PDE down into problems they know how to solve efficiently.

First, their algorithm advances a solution forward in time by solving the heat equation (also called the “diffusion equation”), which models how heat from a source spreads over a shape. Picture using a blow torch to warm up a metal plate — this equation describes how heat from that spot would diffuse over it. 
This step can be completed easily with linear algebra.

Now, imagine that the parabolic PDE has additional nonlinear behaviors that are not described by the spread of heat. This is where the second step of the algorithm comes in: it accounts for the nonlinear piece by solving a Hamilton-Jacobi (HJ) equation, a first-order nonlinear PDE.

While generic HJ equations can be hard to solve, Mattos Da Silva and coauthors prove that their splitting method applied to many important PDEs yields an HJ equation that can be solved via convex optimization algorithms. Convex optimization is a standard tool for which researchers in geometry processing already have efficient and reliable software. In the final step, the algorithm advances a solution forward in time using the heat equation again to advance the more complex second-order parabolic PDE forward in time.

Among other applications, the framework could help simulate fire and flames more efficiently. “There’s a huge pipeline that creates a video with flames being simulated, but at the heart of it is a PDE solver,” says Mattos Da Silva. For these pipelines, an essential step is solving the G-equation, a nonlinear parabolic PDE that models the front propagation of the flame and can be solved using the researchers’ framework.

The team’s algorithm can also solve the diffusion equation in the logarithmic domain, where it becomes nonlinear. Senior author Justin Solomon, associate professor of EECS and leader of the CSAIL Geometric Data Processing Group, previously developed a state-of-the-art technique for optimal transport that requires taking the logarithm of the result of heat diffusion. Mattos Da Silva’s framework provided more reliable computations by doing diffusion directly in the logarithmic domain. This enabled a more stable way to, for example, find a geometric notion of average among distributions on surface meshes like a model of a koala.

Even though their framework focuses on general, nonlinear problems, it can also be used to solve linear PDE. For instance, the method solves the Fokker-Planck equation, where heat diffuses in a linear way, but there are additional terms that drift in the same direction heat is spreading. In a straightforward application, the approach modeled how swirls would evolve over the surface of a triangulated sphere. The result resembles purple-and-brown latte art.

The researchers note that this project is a starting point for tackling the nonlinearity in other PDEs that appear in graphics and geometry processing head-on. For example, they focused on static surfaces but would like to apply their work to moving ones, too. Moreover, their framework solves problems involving a single parabolic PDE, but the team would also like to tackle problems involving coupled parabolic PDE. These types of problems arise in biology and chemistry, where the equation describing the evolution of each agent in a mixture, for example, is linked to the others’ equations.

Mattos Da Silva and Solomon wrote the paper with Oded Stein, assistant professor at the University of Southern California’s Viterbi School of Engineering. Their work was supported, in part, by an MIT Schwarzman College of Computing Fellowship funded by Google, a MathWorks Fellowship, the Swiss National Science Foundation, the U.S. Army Research Office, the U.S. Air Force Office of Scientific Research, the U.S. National Science Foundation, MIT-IBM Watson AI Lab, the Toyota-CSAIL Joint Research Center, Adobe Systems, and Google Research.