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
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.
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.
This is allowable because in the original statement of this problem we assumed that [latex]x>0[/latex]. (If [latex]x=0[/latex] then the original equation becomes [latex]0=2[/latex], which is clearly a false statement.)
In this equation, [latex]p\left(x\right)=-\frac{8}{3x}[/latex] and [latex]q\left(x\right)=-\frac{2}{3x}[/latex].
Subtract [latex]y[/latex] from each side and add [latex]4{x}^{2}-5\text{:}[/latex]
[latex]3y^{\prime} -y=4{x}^{2}-5[/latex].
Next divide both sides by [latex]3\text{:}[/latex]
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].
Hint
Multiply both sides by the common denominator, then collect all terms involving [latex]y[/latex] on one side.
Show Solution
[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]
Try It
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
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
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].
Show Solution
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
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
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.
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.
Watch the following video to see the worked solution to Example: A First-Order Linear Initial-Value Problem.
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.