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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?

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

Further reading: Addendum for derivation of integrating factor

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.

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