How to Use The Integrating Factor Method for Differential Equations

The order of a differential equation is determined by the highest order derivative involved! Therefore, first order differential equations consist only of a function and its derivative.

A first order differential equation can be written as

y^{'} + f (x) y = g (x)

You solve these types of equations using integrating factors.

Theory

The integrating factor is $e^{F (x)}$ , where

F (x) = \int f (x) d x

and $F^{'} (x) = f (x)$ without $C$ .

By using the product rule, you see that

{(e^{F (x)} y)}^{'} = e^{F (x)} y^{'} + f (x) e^{F (x)} y

This is what you want to introduce into the original equation. Then the differential equation can be solved by integrating factors.

Rule

\begin{array}{l} y^{'} + f (x) y & = g (x) | \cdot e^{F (x)} \\ e^{F (x)} y^{'} + f (x) e^{F (x)} y & = {(e^{F (x)} y)}^{'} = e^{F (x)} g (x) \\ \int {(e^{F (x)} y)}^{'} d x & = \int e^{F (x)} g (x) d x \\ e^{F (x)} y & = \int e^{F (x)} g (x) d x \\ y & = \int e^{F (x)} g (x) d x \\ \cdot e^{- F (x)} \end{array}

\begin{array}{l} y^{'} + f (x) y & = g (x) | \cdot e^{F (x)} \\ e^{F (x)} y^{'} + f (x) e^{F (x)} y & = {(e^{F (x)} y)}^{'} = e^{F (x)} g (x) \\ \int {(e^{F (x)} y)}^{'} d x & = \int e^{F (x)} g (x) d x \\ e^{F (x)} y & = \int e^{F (x)} g (x) d x \\ y & = \int e^{F (x)} g (x) d x \cdot e^{- F (x)} \end{array}

Example 1

Solve the differential equation $y^{'} + 2 x y = e^{x}$ by using integrating factors

\begin{array}{l} y^{'} + 2 x y & = e^{x} & | \cdot e^{x^{2}} \\ e^{x^{2}} y^{'} + 2 x e^{x^{2}} y & = 2 x e^{x^{2}} \\ {(e^{x^{2}} y)}^{'} & = 2 x e^{x^{2}} \\ \int {(e^{x^{2}} y)}^{'} d x & = \int 2 x e^{x^{2}} d x \\ e^{x^{2}} y & = \int 2 x e^{x^{2}} d x \\ e^{x^{2}} y & \overset{*}{=} \int 2x e^{u} \frac{1}{2x} d u \\ e^{x^{2}} y & = \int e^{u} d u \\ e^{x^{2}} y & = e^{u} + C \\ e^{x^{2}} y & = e^{x^{2}} + C & | \cdot e^{- x^{2}} \\ y & = 1 + C e^{- x^{2}} \end{array}

\begin{array}{l} y^{'} + 2 x y & = e^{x} & | \cdot e^{x^{2}} \\ e^{x^{2}} y^{'} + 2 x e^{x^{2}} y & = 2 x e^{x^{2}} \\ {(e^{x^{2}} y)}^{'} & = 2 x e^{x^{2}} \\ \int {(e^{x^{2}} y)}^{'} d x & = \int 2 x e^{x^{2}} d x \\ e^{x^{2}} y & = \int 2 x e^{x^{2}} d x \\ e^{x^{2}} y & \overset{*}{=} \int 2x e^{u} \frac{1}{2x} d u \\ e^{x^{2}} y & = \int e^{u} d u \\ e^{x^{2}} y & = e^{u} + C \\ e^{x^{2}} y & = e^{x^{2}} + C & | \cdot e^{- x^{2}} \\ y & = 1 + C e^{- x^{2}} \end{array}

\begin{aligned} u & = x^{2} \\ \frac{d u}{d x} & = 2 x \\ d u & = 2 x d x \\ \frac{1}{2 x} d u & = d x \end{aligned}

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