If the solution curve $y=f(x)$ of the differential equation $(x^{2}-4)y^{\prime}-2xy+2x(4-x^{2})^{2}=0$ for $x>2$ passes through the point $(3, 15)$, then the local maximum value of $f$ is:

Let $y=y(x)$ be the solution of the differential equation $(1-x^2) dy = [xy + (x^3+2) \sqrt{3(1-x^2)}] dx$ for $-1 < x < 1$,with $y(0)=0$. If $y(1/2) = m/n$,where $m$ and $n$ are coprime numbers,then $m+n$ is equal to . . . . . . . . . .

The solution of the differential equation $\left( {{e^{{x^2}}} + {e^{{y^2}}}} \right) y \frac{{dy}}{{dx}} + {e^{{x^2}}}(x{y^2} - x) = 0$ is

If the solution of the differential equation $\frac{dy}{dx} + e^x(x^2 - 2)y = (x^2 - 2x)(x^2 - 2)e^{2x}$ satisfies $y(0) = 0$,then the value of $y(2)$ is

If $\frac{dx}{dy} = \frac{1+x-y^2}{y}$ and $x(1) = 1$,then $5x(2)$ is equal to:

The solution of the differential equation $(1+x) \frac{dy}{dx} - xy = 1-x$ is