The area (in sq. units) of the region bounded by the curves $y=x^2$ and $y=8-x^2$ is
$\frac{32}{3}$
$\frac{16}{3}$
$\frac{64}{3}$
$\frac{128}{3}$
The solution of the differential equation $x^2(y+1) \frac{d y}{d x}+y^2(x+1)^2=0$, when $y(1)=2$, is
$\log \left|x^2 y\right|=\frac{2}{x}+\frac{1}{y}+x-1$
$\log \left|\frac{1}{4} x^2 y\right|=\frac{1}{x}+\frac{2}{y}+x-1$
$\log \left|\frac{1}{2} x^2 y\right|=\frac{1}{x}+\frac{1}{y}-x-\frac{1}{2}$
$\log \left|\frac{1}{3} x^2 y\right|=\frac{1}{x}+\frac{1}{y}-x+\frac{1}{2}$
The general solution of the differential equation $\frac{d y}{d x}=\frac{2 x+y-3}{2 y-x+3}$
$x^2-x y-y^2+3 x+3 y+c=0$
$x^2-x y-y^2-3 x-3 y+c=0$
$x^2+x y-y^2-3 x-3 y+c=0$
$x^2+x y+y^2+3 x-3 y+c=0$
If $x \log x \frac{d y}{d x}+y=\log x^2$ and $y(e)=0$, then $y\left(e^2\right)=$
0
1
$\frac{1}{2}$
$\frac{3}{2}$
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