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Question Number 75175 by Rio Michael last updated on 08/Dec/19

$$solvethedifferentialequation$$$$(x^2-1)\frac{dy}{dx}+2y=0wheny=3andx=2,expressing$$$$youranswerintheformy=f\left(x\right)$$

Answered by Kunal12588 last updated on 08/Dec/19

$$(x^2-1)\frac{dy}{dx}+2y=0$$$$\Rightarrow \frac{dy}{2y}=-\frac{dx}{x^2-1}$$$$\Rightarrow \frac{1}{2}\int \frac{dy}{y}=\int \frac{dx}{1-x^2}$$$$\frac{1}{1-x^2}=\frac{A}{(1-x)}+\frac{B}{(1+x)}$$$$\Rightarrow 1=A(1+x)+B(1-x)$$$$\Rightarrow A=\frac{1}{2},B=\frac{1}{2}$$$$\frac{1}{2}log\left(y\right)=\frac{1}{2}\int \frac{dx}{1-x}+\frac{1}{2}\int \frac{dx}{1+x}$$$$\Rightarrow logy=-log(1-x)+log(1+x)+logC$$$$\Rightarrow y=\frac{(1+x)C}{1-x}$$$$3=\frac{3C}{-1}$$$$\Rightarrow C=-1$$$$y=\frac{x+1}{x-1}$$

Commented by Rio Michael last updated on 08/Dec/19

$$thankyousir$$

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