Q

Find the General solution. (1 + x ^2) dy + 2xy dx = cot x dx ( x not equal to 0)

Find the general solution.

Q8.    $(1 + x^2)dy + 2xydx = \cot x dx\ (x\neq 0)$

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Given equation is
$(1 + x^2)dy + 2xydx = \cot x dx\ (x\neq 0)$
we can rewrite it as
$\frac{dy}{dx}+\frac{2xy}{(1+x^2)}= \frac{\cot x}{1+x^2}$
This is  $\frac{dy}{dx} + py = Q$  type where $p = \frac{2x}{1+ x^2}$ and $Q =\frac{\cot x}{1+x^2}$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{2x}{1+ x^2} dx}= e^{\log(1+ x^2)} = 1+x^2$
Now, the solution of the given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(1+x^2) =\int ((\frac{\cot x}{1+x^2})\times (1+ x^2))dx +C$
$y(1+x^2) =\int \cot x dx+C\\ \\ y(1+x^2)= \log |\sin x|+ C\\ \\ y = (1+x^2)^{-1}\log |\sin x|+C(1+x^2)^{-1}$
Therefore, the general solution is   $y = (1+x^2)^{-1}\log |\sin x|+C(1+x^2)^{-1}$

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