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

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

User Shibbir
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1 Answer

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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)}

User Cleve Green
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