The general solution of left(left(1+x^2right) | vedclass

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(B) Given differential equation is $\left(\left(1+x^2\right) y \sin x-2 x y\right) d x-\log y^{1+x^2} d y=0$. Dividing by $y$ and using $\log y^{1+x^2

Vedclass · Accepted Answer

$(\log y)^2+2 \cos x+\log \left(1+x^2\right)^2=c$

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(B) Given differential equation is $\left(\left(1+x^2\right) y \sin x-2 x y\right) d x-\log y^{1+x^2} d y=0$. Dividing by $y$ and using $\log y^{1+x^2} = (1+x^2) \log y$, we get: $\left((1+x^2) \sin x - 2x\right) dx - (1+x^2) \log y \frac{dy}{y} = 0$. Rearranging the terms: $\left(\sin x - \frac{2x}{1+x^2}\right) dx = \frac{\log y}{y} dy$. Integrating both sides: $\int \left(\sin x - \frac{2x}{1+x^2}\right) dx = \int \frac{\log y}{y} dy$. Let $u = \log y$, then $du = \frac{1}{y} dy$. Integrating gives: $-\cos x - \log(1+x^2) = \frac{(\log y)^2}{2} + C_1$. Multiplying by $2$: $-2 \cos x - 2 \log(1+x^2) = (\log y)^2 + 2C_1$. Rearranging: $(\log y)^2 + 2 \cos x + 2 \log(1+x^2) = C$. Since $2 \log(1+x^2) = \log(1+x^2)^2$, the solution is $(\log y)^2 + 2 \cos x + \log(1+x^2)^2 = C$.

The general solution of $\left(\left(1+x^2\right) y \sin x-2 x y\right) d x-\log y^{1+x^2} d y=0$ is

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