First-Order Linear Equations and Applications: Learn It 3

Giselle Miller

First-Order Linear Equations and Applications: Learn It 3

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:

[latex]y^{\prime} +p\left(x\right)y=q\left(x\right)[/latex].

The first term on the left-hand side 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 product rule from differentiation.

If we multiply [latex]y' +p(x)y=q(x)[/latex] by a yet-to-be-determined function [latex]\mu(x)[/latex], then the equation becomes:

[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].

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.

Recall the product rule:

[latex]\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)[/latex]

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

[latex]{\mu }^{\prime }\left(x\right)=\mu \left(x\right)p\left(x\right)[/latex].

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

[latex]\begin{array}{ccc}\hfill \frac{{\mu }^{\prime }\left(x\right)}{\mu \left(x\right)}& =\hfill & p\left(x\right)\hfill \\ \hfill {\displaystyle\int \frac{{\mu }^{\prime }\left(x\right)}{\mu \left(x\right)}dx}& =\hfill & {\displaystyle\int p\left(x\right)dx}\hfill \\ \hfill \text{ln}|\mu \left(x\right)|& =\hfill & {\displaystyle\int p\left(x\right)dx+C}\hfill \\ \hfill {e}^{\text{ln}|\mu \left(x\right)|}& =\hfill & {e}^{\displaystyle\int p\left(x\right)dx+C}\hfill \\ \hfill |\mu \left(x\right)|& =\hfill & {C}_{1}{e}^{\displaystyle\int p\left(x\right)dx}\hfill \\ \hfill \mu \left(x\right)& =\hfill & {C}_{2}{e}^{\displaystyle\int p\left(x\right)dx}\hfill \end{array}[/latex]

Here [latex]C_2[/latex] can be an arbitrary (positive or negative) constant. Since we only need one integrating factor to solve our equation, we can set [latex]C_2 = 1[/latex]. This leads to a general method for solving a first-order linear differential equation.

We first multiply both sides of [latex]y' +p(x)y=q(x)[/latex] by the integrating factor [latex]\mu(x)[/latex]. This gives:

[latex]\mu(x)y' +\mu(x)p(x)y=\mu(x)q(x)[/latex]

The left-hand side can be rewritten as [latex]\frac{d}{dx}(\mu(x)y)[/latex]:

[latex]\frac{d}{dx}(\mu(x)y)=\mu(x)q(x)[/latex]

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

[latex]\begin{array}{ccc}\hfill \int \frac{d}{dx}(\mu(x)y)dx& =\hfill & \int \mu(x)q(x)dx\hfill \ \hfill \mu(x)y& =\hfill & \int \mu(x)q(x)dx\hfill \end{array}[/latex]

Divide both sides by [latex]\mu(x)[/latex]:

[latex]y=\frac{1}{\mu(x)}\left[\int \mu(x)q(x)dx+C\right][/latex]

Since [latex]\mu(x)[/latex] was previously calculated, we are now finished.

integrating factor

An integrating factor is a function [latex]\mu(x)[/latex] that, when multiplied by a first-order linear differential equation, makes the left side equal to the derivative of a product. For the standard form [latex]y' + p(x)y = q(x)[/latex], the integrating factor is:

[latex]\mu(x) = e^{\int p(x)dx}[/latex]

When finding the integrating factor, we can set the constant of integration to zero since we only need one specific integrating factor. However, when solving for [latex]y[/latex], we must include the constant [latex]C[/latex] to get the general family of solutions.

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].

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

To put this differential equation into standard form, divide both sides by [latex]x\text{:}[/latex]

[latex]y^{\prime} +\frac{3}{x}y=4x - 3[/latex].

Therefore [latex]p\left(x\right)=\frac{3}{x}[/latex] and [latex]q\left(x\right)=4x - 3[/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].
Multiplying both sides of the differential equation by [latex]\mu \left(x\right)[/latex] gives us

[latex]\begin{array}{ccc}\hfill {x}^{3}{y}^{\prime }+{x}^{3}\left(\frac{3}{x}\right)y& =\hfill & {x}^{3}\left(4x - 3\right)\hfill \\ \hfill {x}^{3}{y}^{\prime }+3{x}^{2}y& =\hfill & 4{x}^{4}-3{x}^{3}\hfill \\ \hfill \frac{d}{dx}\left({x}^{3}y\right)& =\hfill & 4{x}^{4}-3{x}^{3}.\hfill \end{array}[/latex]
Integrate both sides of the equation.

[latex]\begin{array}{ccc}\hfill {\displaystyle\int \frac{d}{dx}\left({x}^{3}y\right)dx}& =\hfill & {\displaystyle\int 4{x}^{4}-3{x}^{3}dx}\hfill \\ \hfill {x}^{3}y& =\hfill & \frac{4{x}^{5}}{5}-\frac{3{x}^{4}}{4}+C\hfill \\ \hfill y& =\hfill & \frac{4{x}^{2}}{5}-\frac{3x}{4}+C{x}^{-3}.\hfill \end{array}[/latex]
There is no initial value, so the problem is complete.

Analysis

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

[latex]{e}^{\displaystyle\int p\left(x\right)dx=}{e}^{\displaystyle\int \left(\frac{3}{x}\right)dx}={e}^{3\text{ln}|x|}={|x|}^{3}[/latex],

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 the example above.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.You can view the transcript for this segmented clip of “4.5.2” here (opens in new window).

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

Solve the initial-value problem

[latex]{y}^{\prime }+3y=2x - 1,y\left(0\right)=3[/latex].

Watch the following video to see the worked solution to the example above.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.You can view the transcript for this segmented clip of “4.5.2” here (opens in new window).