Write differential equation in standard form

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]^<\prime >+\fracy=\frac[/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]^<\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

[latex]\mu \left(x\right)^<\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:

Matching term by term gives [latex]y=f\left(x\right),g\left(x\right)=\mu \left(x\right)[/latex], and [latex]^<\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]^<\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]_[/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]^<\prime >+p\left(x\right)y=q\left(x\right)[/latex] by the integrating factor [latex]\mu \left(x\right)[/latex]. This gives

[latex]\mu \left(x\right)^<\prime >+\mu \left(x\right)p\left(x\right)y=\mu \left(x\right)q\left(x\right)[/latex].

The left-hand side of the above equation can be rewritten as [latex]\frac\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]^<\prime >+\fracy=\frac[/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\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]^<\prime >+\fracy=\frac[/latex]. This integrating factor guarantees just that.

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

1. Put the equation into standard form and identify [latex]p\left(x\right)[/latex] and [latex]q\left(x\right)[/latex].
2. Calculate the integrating factor [latex]\mu \left(x\right)=^<\displaystyle\int p\left(x\right)dx>[/latex].
3. Multiply both sides of the differential equation by [latex]\mu \left(x\right)[/latex].
4. Integrate both sides of the equation obtained in step [latex]3[/latex], and divide both sides by [latex]\mu \left(x\right)[/latex].
5. 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^-3x[/latex]. Assume [latex]x>0[/latex].

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

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)^<\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)=^<\displaystyle\int p\left(x\right)dx>[/latex]. For this [latex]p\left(x\right)[/latex] we get

^<\displaystyle\int p\left(x\right)dx=>^<\displaystyle\int \left(\frac\right)dx>=^|x|>=<|x|>^[/latex],

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

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try it

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

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