Solve the differential equation (x^2-1) dy dx +2xy=1 x^2-1 .

Question

Solve the differential equation $(\text{x}^2-1)\frac{\text{dy}}{\text{dx}}+2\text{xy}=\frac{1}{\text{x}^2-1}.$

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Answer

Given differential equation is
$(\text{x}^2-1)\frac{\text{dy}}{\text{dx}}+2\text{xy}=\frac{1}{\text{x}^2-1}$
divide on both sides by $(\text{x}^2-1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\Big(\frac{2\text{x}}{\text{x}^2-1}\Big)\text{y}=\frac{1}{(\text{x}^2-1)^2}$
Which is a linear differential equation.
On Comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get
$\text{P}=\frac{2\text{x}}{\text{x}^2-1},\text{Q}=\frac{1}{(\text{x}^2-1)^2}$
$\text{I.F.}=\text{e}^{\int\text{pdx}}$
$\text{I.F.}=\text{e}^{\int{\big(\frac{2\text{x}}{\text{x}^2-1}\big)}\text{dx}}$
Put $\text{x}^2-1=\text{t}$
$\Rightarrow2\text{xdx}=\text{dt}$
$\therefore\text{I.F.}=\text{e}^{\int\frac{\text{dt}}{\text{t}}}=\text{e}^{\log\text{t}}$
$\text{I.F.}=\text{t}=(\text{x}^2-1)$
The complete solution is
$\text{y}\text{I.F.}=\int\text{Q}\text{I.F.}+\text{k}$
$\Rightarrow\text{y.}(\text{x}^2-1)=\int\frac{1}{(\text{x}^2-1)^2}.(\text{x}^2-1)\text{dx}+\text{k}$
$\Rightarrow\text{y.}(\text{x}^2-1)=\int\frac{\text{dx}}{(\text{x}^2-1)}$
$\Rightarrow\text{y.}(\text{x}^2-1)=\frac{1}{2}\log\Big(\frac{\text{x}-1}{\text{x}+1}\Big)+\text{k}$

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