$$ \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x= $$
$\frac{\pi^2}{4}$
$\frac{\pi}{2}$
$\frac{\pi^2}{2}$
$\frac{\pi}{4}$
The differential equation corresponding to the family of parabolas whose axis is along $x=1$ is
$\frac{d^2 y}{d x^2}-(x-1) \frac{d y}{d x}=0$
$(x-1) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$
$\frac{d^2 y}{d x^2}+(x-1) \frac{d y}{d x}-y=0$
$(x-1) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0$
The general solution of the equation $\frac{d y}{d x}+\frac{1}{x} y=\frac{1}{x} e^x$
$y=x e^x+c$
$y=x e^x+c e^{-x}$
$y=\frac{e^x+c}{x}$
$y=\frac{e^{-x}+c x}{x}$
The general solution of the differential equation
$$ \left(x \sin \frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x $$
$\log x+\tan \frac{y}{x}=C$
$\log x+\cos \frac{y}{x}=C$
$\log x-\sin \frac{y}{x}=C$
$\log x-\cos \frac{y}{x}=C$
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