How to Solve Linear First Order Differential Equations

Last Updated: June 17, 2017

A linear first order ordinary differential equation is that of the following form, where we consider that $y=y(x),$ and $y$ and its derivative are both of the first degree.

${\frac {{\mathrm {d}}y}{{\mathrm {d}}x}}+P(x)y=Q(x)$

To solve this equation, we use an integrating factor $e^{{\int P(x){\mathrm {d}}x}}.$ We will provide an example and show that this integrating factor makes the above equation exact, as intended.

Part 1

Part 1 of 2:

Example

1
Solve the following equation. Because the degree of $y$ and its derivative are both 1, this equation is linear.
- ${\frac {{\mathrm {d}}y}{{\mathrm {d}}x}}+{\frac {2}{x}}y=4\sin x$
2
Find the integrating factor.
- ${\begin{aligned}e^{{\int P(x){\mathrm {d}}x}}&=e^{{\int 2/x{\mathrm {d}}x}}\\&=e^{{2\ln |x|}}\\&=x^{{2}}\end{aligned}}$
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3
Rewrite the equation in Pfaffian form and multiply by the integrating factor. We can confirm that this is an exact differential equation by doing the partial derivatives.
- $(2xy-4x^{{2}}\sin x){\mathrm {d}}x+x^{{2}}{\mathrm {d}}y=0$
4
Solve this equation using any means possible. We write $f$ as a solution to the differential equation.
- $f=\int Q{\mathrm {d}}y=x^{{2}}y+R(x)$
- ${\frac {\partial f}{\partial x}}=P=2xy+{\frac {{\mathrm {d}}R}{{\mathrm {d}}x}}$
- $R(x)=(4x^{{2}}-8)\cos x-8x\sin x$
- $x^{{2}}y+(4x^{{2}}-8)\cos x-8x\sin x=C$
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Part 2

Part 2 of 2:

Derivation of the Integrating Factor

1
Rewrite the linear differential equation in Pfaffian form.
- $(P(x)y-Q(x)){\mathrm {d}}x+{\mathrm {d}}y=0$
2
Consider an integrating factor $\mu (x)$ . This integrating factor is such that multiplying the above equation by it makes the equation exact.
- $(\mu P(x)y-\mu Q(x)){\mathrm {d}}x+\mu (x){\mathrm {d}}y=0$
3
Invoke the necessary and sufficient condition for exactness. To be exact, the coefficients of the differentials must satisfy Clariaut's theorem.
- ${\frac {\partial }{\partial y}}(\mu P(x)y-\mu Q(x))={\frac {\partial \mu }{\partial x}}$
4
Simplify the resulting expression. We recognize that $P(x),$ $Q(x),$ and $\mu$ are all functions of $x$ only.
- $\mu P(x)={\frac {{\mathrm {d}}\mu }{{\mathrm {d}}x}}$
5
Separate variables and integrate to solve for $\mu$ .
- ${\begin{aligned}{\frac {1}{\mu }}{\mathrm {d}}\mu &=P(x){\mathrm {d}}x\\\ln |\mu |&=\int P(x){\mathrm {d}}x\\\mu &=e^{{\int P(x){\mathrm {d}}x}}\ \ \ \ \ \ ({\text{QED.}})\end{aligned}}$
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