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 $y(x)$.

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

The equation of the curve passing through the origin and satisfying the differential equation $(1+x^2) \frac{dy}{dx} + 2xy = 4x^2$ is:

Let $y = y(x)$ be the solution of the differential equation $\cos x \frac{dy}{dx} + 2y \sin x = \sin 2x$ for $x \in (0, \frac{\pi}{2})$. If $y(\frac{\pi}{3}) = 0$,then $y(\frac{\pi}{4})$ is equal to:

Integrating factor of $(x+2 y^3) \frac{d y}{d x}=y^2$ is

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 3(\tan^2 x + 1)y = \sec^2 x$,with the initial condition $y(0) = \frac{1}{3} + e^3$. Then $y\left(\frac{\pi}{4}\right)$ is equal to: