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(A) The given differential equation is $\left(1+x^{2}\right) \frac{dy}{dx}+2 xy=\frac{1}{1+x^{2}}$.Dividing by $\left(1+x^{2}\right)$,we get $\frac{dy

Vedclass · Accepted Answer

$y\left(1+x^{2}ight)=	an ^{-1} x-\frac{\pi}{4}$

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(A) The given differential equation is $\left(1+x^{2}ight) \frac{dy}{dx}+2 xy=\frac{1}{1+x^{2}}$.Dividing by $\left(1+x^{2}ight)$,we get $\frac{dy}{dx}+\frac{2x}{1+x^{2}}y=\frac{1}{(1+x^{2})^{2}}$.This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=\frac{2x}{1+x^{2}}$ and $Q=\frac{1}{(1+x^{2})^{2}}$.The integrating factor $I.F. = e^{\int P dx} = e^{\int \frac{2x}{1+x^{2}} dx} = e^{\ln(1+x^{2})} = 1+x^{2}$.The general solution is $y(I.F.) = \int (Q 	imes I.F.) dx + C$.$y(1+x^{2}) = \int \left(\frac{1}{(1+x^{2})^{2}} \cdot (1+x^{2})ight) dx + C$.$y(1+x^{2}) = \int \frac{1}{1+x^{2}} dx + C$.$y(1+x^{2}) = 	an^{-1} x + C$  --- $(1)$.Given $y=0$ when $x=1$,we substitute these values into $(1)$:$0(1+1^{2}) = 	an^{-1}(1) + C \Rightarrow 0 = \frac{\pi}{4} + C \Rightarrow C = -\frac{\pi}{4}$.Substituting $C$ back into $(1)$,the particular solution is $y(1+x^{2}) = 	an^{-1} x - \frac{\pi}{4}$.

Find a particular solution satisfying the given condition: $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^{2}}$; $y=0$ when $x=1$.

Similar Questions

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