# Integrating Factor Usage

Sometimes it can be difficult deciding which method to use to solve differential equations. Can the integrating factor method always be used when solving differential equations?

• No, that's why a variety of methods exist. – Teh Rod Jul 9 '17 at 0:55

The usage of integrating factor is to find a solution to differential equation. Integrating factor is used when we have the following first order linear differential equation. It can be homogeneous(when $Q(x)=0$) or non homogeneous.

$$\frac{dy}{dx}+P(x)y=Q(x)$$

where $P(x)$ & $Q(x)$ is a function of $x$.

The integrating factor is then,

$$\mu=e^{\int P(x)dx}$$

Suppose we have the following standard differential

$$M(x,y)dx+N(x,y)dy=0$$

Further suppose that the above equation is not exact at all!

$$\frac{\partial M(x,y)}{\partial y}\neq\frac{\partial N(x,y)}{\partial x}$$

There exists an integrating factor $\mu$ that will make this equation exact such that when we multiply it into the differential

$$\mu(x)M(x,y)dx+\mu(x)N(x,y)dy=0$$

$$\frac{\partial \mu(x)M(x,y)}{\partial y}=\frac{\partial \mu(x)N(x,y)}{\partial x}$$

However, multiplying an integrating factor might cause a gain of new solution or loss of original solution.

Integrating factor method can only be used when there exists an integrating factor.

To answer your question: 'Can the integrating factor method always be used when solving differential equations'? The answer will be not always. When an integrating factor exists please use it.

Suppose that we our integrating factor depends only on x such that

$$\mu(x)M(x,y)dx+\mu(x)N(x,y)dy=0$$

Will make this equation exact!

$$\frac{\partial \mu(x)M(x,y)}{\partial y}=\frac{\partial \mu(x)N(x,y)}{\partial x}$$

Notice that the integrating factor depends only x and is independent of y.

$$\mu(x)\frac{\partial M(x,y)}{\partial y}=\mu(x)\frac{\partial N(x,y)}{\partial x}+N(x,y)\frac{\partial \mu(x)}{\partial x}$$

$$\mu(x)\frac{\partial M(x,y)}{\partial y}-\mu(x)\frac{\partial N(x,y)}{\partial x}=N(x,y)\frac{d\mu(x)}{dx}$$

$$\frac{1}{N(x,y)}\left[\mu(x)\frac{\partial M(x,y)}{\partial y}-\mu(x)\frac{\partial N(x,y)}{\partial x}\right]=\frac{d\mu(x)}{dx}$$

$$\int \frac{1}{N(x,y)}\left[\frac{\partial M(x,y)}{\partial y}-\frac{\partial N(x,y)}{\partial x}\right]dx=\int\frac{d\mu(x)}{\mu(x)}$$

$$\mu (x)=exp\int \frac{1}{N(x,y)}\left[\frac{\partial M(x,y)}{\partial y}-\frac{\partial N(x,y)}{\partial x}\right]dx$$

If the the integrating factor depends on y then you can try it yourself. Similar method of solving it.