Let $f:[2,5] \rightarrow R$ be a differentiable function and $\frac{f(5)}{f(2)}=1$. If there is a $c \in (2,5)$ such that $c f^{\prime}(c)=2 f(c)-2 c^3$, then $f(x)=$

If $x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x$,then $x^3 y^{-3}$ is equal to

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

For all $x > 0$,let $y_1(x), y_2(x)$,and $y_3(x)$ be the functions satisfying $\frac{dy_1}{dx} - (\sin x)^2 y_1 = 0, y_1(1) = 5$; $\frac{dy_2}{dx} - (\cos x)^2 y_2 = 0, y_2(1) = \frac{1}{3}$; and $\frac{dy_3}{dx} - \left(\frac{2-x^3}{x^3}\right) y_3 = 0, y_3(1) = \frac{3}{5e}$ respectively. Then $\lim_{x \rightarrow 0^{+}} \frac{y_1(x) y_2(x) y_3(x) + 2x}{e^{3x} \sin x}$ is equal to:

Let $f$ be a differentiable function $f : R \rightarrow R$ satisfying the equation $f(x) = (1+x^2) \left[ 1 + \int_{0}^{x} \frac{f(t)}{1+t^2} dt \right]$ for all $x \in R$. Then $f(1)$ is:

If $y=y(x)$ is the solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ satisfying $y(1) = 1$,then the value of $y\left(\frac{1}{2}\right)$ is