Question

Give the largest interval I over which the general solution is defined. PLEASE EXPLAIN HOW!!!

(x^2-1)dy/dx+2y=(x+1)^2

Answer 1

Divide both sides by $x^2-1$ to get a linear ODE,

$\dfrac{\mathrm dy}{\mathrm dx}+\dfrac2{x^2-1}y=\dfrac{x+1}{x-1}$

In order for this operation to be valid in the first place, we require that $x\neq\pm1$ (since that would make $\dfrac1{x^2-1}$ undefined, which we don't want to happen). Then we are forcing any solution to the ODE to exist on any of the three intervals, $(-\infty,-1)$, $(-1, 1)$, or $(1,\infty)$, and either the first or third of these can be chosen as the largest interval.

In case you also need to solve the ODE: Multiply both sides by $\dfrac{1-x}{1+x}$, so that

$\dfrac{1-x}{1+x}\dfrac{\mathrm dy}{\mathrm dx}-\dfrac2{(1+x)^2}y=-1$

Then the left side can be condensed as the derivative of a product, since

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{1-x}{1+x}\right]=-\dfrac2{(1+x)^2}$

and we have

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{1-x}{1+x}y\right]=-1$

Integrate both sides:

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\left[\frac{1-x}{1+x}y\right]\,\mathrm dx=-\int\mathrm dx$

$\dfrac{1-x}{1+x}y=-x+C$

$\implies\boxed{y=\dfrac{(-x+C)(1+x)}{1-x}}$