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,

`(\mathrm dy)/(\mathrm dx)+\frac2{x^2-1}y=(x+1)/(x-1)`

In order for this operation to be valid in the first place, we require that
`x\\eq\pm1` (since that would make
`\frac1{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
`(1-x)/(1+x)`, so that

`(1-x)/(1+x)(\mathrm dy)/(\mathrm dx)-\frac2{(1+x)^2}y=-1`

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

`(\mathrm d)/(\mathrm dx)\left[(1-x)/(1+x)\right]=-\frac2{(1+x)^2}`

and we have

`(\mathrm d)/(\mathrm dx)\left[(1-x)/(1+x)y\right]=-1`

Integrate both sides:

`\displaystyle\int(\mathrm d)/(\mathrm dx)\left[(1-x)/(1+x)y\right]\,\mathrm dx=-\int\mathrm dx`

`(1-x)/(1+x)y=-x+C`

`\implies\boxed{y=((-x+C)(1+x))/(1-x)}`