If the function $y = f(x)$ satisfies the differential equation $(x^3 + 1)dy = x(1 - 3xy)dx$ and $f(0) = 0$,then $\mathop {\lim }\limits_{x \to 0} \frac{x^2}{f(x)}$ is equal to

The solution of the differential equation $2 \frac{dy}{dx} - \frac{y}{x} = \frac{y^2}{x^2}$,given that $y = 2$ when $x = 1$,is

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx} + \frac{\sqrt{2}y}{2\cos^4 x - \cos 2x} = x e^{\tan^{-1}(\sqrt{2} \cot 2x)}$,$0 < x < \pi/2$ with $y(\pi/4) = \pi^2/32$. If $y(\pi/3) = \frac{\pi^2}{18} e^{-\tan^{-1}(\alpha)}$,then the value of $3\alpha^2$ is equal to

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=2(y+2 \sin x-5)x-2 \cos x$ such that $y(0)=7$. Then $y(\pi)$ is equal to :

If $y=y(x)$ is the solution curve of the differential equation $x^{2} dy + (y - \frac{1}{x}) dx = 0$ for $x > 0$ and $y(1) = 1$,then $y(\frac{1}{2})$ is equal to:

If $x y^{\prime}+y-e^x=0$ and $y(a)=b$, then $\lim _{x \rightarrow 1} y(x)$ is