Solving a ODE with integrating factor

Question

$$(x^2 -1) \frac{dy}{dx} + 2xy = x$$

First I divided both sides by $(x^2 - 1)$ $$ \frac{dy}{dx} + {2x \over x^2 - 1}y = {x \over x^2 - 1} $$

and then I found the integrating factor $x^2 - 1$ and multiples both sides with this and that leads me to the initial question... How do I solve this? Thanks

Answer 1

The fact that the integrating factor lead you right back to the original equation should show you that it's already in the form of the product rule. Observe that

$$ (x^2-1)\frac{dy}{dx} + 2xy = \frac{d}{dx}\big((x^2-1)y\big) $$

Therefore

$$ \frac{d}{dx}\big((x^2-1)y\big) = x $$

And you can integrate throughout to get

$$ (x^2-1)y = \frac{x^2}{2} + c $$

Does this make sense?

Answer 2

Now you know that you can get some insight into the problem by computing $$ \frac{d}{dx}\Bigl((x^2-1)y(x)\Bigr). $$ The idea of an integrating factor $\mu$ is that you can go from $y'+py=q$ via $\frac{μ'}μ=p$ to $$ (μy)'=μy'+μ'y=μq\implies y=\frac1μ\int μq\,dx. $$ That is, the solution of the ODE is reduced to the computation of an "ordinary" integral.

Answer 3

This equation dosen't need an integrating factor. Its already a derivative $$(x^2 -1) \frac{dy}{dx} + 2xy = x$$ $$\mu (x)y'+\mu ' (x)y=x$$ where $\mu =x^2-1$ $$(\mu (x)y)'=x$$ Integrate both side to get the function y(x)