For $x \in R$,let $y(x)$ be a solution of the differential equation $(x^2-5) \frac{dy}{dx} - 2xy = -2x(x^2-5)^2$ such that $y(2)=7$. Find the maximum value of $y(x)$.

Let $x=x(y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x+\left(y^{2}-4 x e^{x / y^{2}}\right) d y=0$ such that $x(1)=0$. Then,$x(e)$ is equal to

Let $y=y(x)$ be the solution curve of the differential equation $\sin(2x^2) \ln(\tan x^2) dy + (4xy - 4\sqrt{2}x \sin(x^2 - \frac{\pi}{4})) dx = 0$ for $0 < x < \sqrt{\frac{\pi}{2}}$,which passes through the point $(\sqrt{\frac{\pi}{6}}, 1)$. Then $|y(\sqrt{\frac{\pi}{3}})|$ is equal to $.....$

Let $y=y(x)$ be a solution curve of the differential equation $(y+1) \tan ^{2} x \,dx+\tan x \,dy+y \,dx=0$ for $x \in \left(0, \frac{\pi}{2}\right)$. If $\lim _{x \rightarrow 0+} x y(x)=1$,then the value of $y\left(\frac{\pi}{4}\right)$ is:

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x$,$x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$,such that $y(0) = 1$. Then

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