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?
1
-
1$\begingroup$ No, that's why a variety of methods exist. $\endgroup$– Teh RodCommented Jul 9, 2017 at 0:55
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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.
