Let us consider a curve $y=f(x)$ passing through the point $(-2, 2)$ and the slope of the tangent to the curve at any point $(x, f(x))$ is given by $f(x)+x f'(x)=x^2$. Then:

The solution of $y' - y = 1, y(0) = -1$ is given by $y(x) = $

If $y=f(x)$ is the solution of the differential equation $x \frac{dy}{dx} = x^2 + 3y$,$x > 0$,$y(2) = 4$,then $f(4) = $ ?

Let $y=y(x)$ be the solution curve of the differential equation $\sec y \frac{dy}{dx} + 2x \sin y = x^3 \cos y$,with the initial condition $y(1) = 0$. Then the value of $y(\sqrt{3})$ is equal to:

Let $F:[3,5] \rightarrow R$ be a twice differentiable function on $(3,5)$ such that $F(x)=e^{-x} \int_{3}^{x} (3t^{2}+2t+4F^{\prime}(t)) \,dt$. If $F^{\prime}(4)=\frac{\alpha e^{\beta}-224}{(e^{\beta}-4)^{2}}$,then $\alpha+\beta$ is equal to $....$

If $\cos x \frac{dy}{dx} = y \sin x - 1$,where $x \neq (2n+1) \frac{\pi}{2}, n \in Z$,is the differential equation corresponding to the curve $y = f(x)$ and $f(0) = 1$,then $f(x) =$