The solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ is:

Let $f : R \rightarrow R$ be a differentiable function with $f(0)=1$ and satisfying the equation $f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y)$ for all $x, y \in R$. Then,the value of $\log _e(f(4))$ is:

Let $f$ be a differentiable function with $\lim_{x \rightarrow \infty} f(x) = 0$. If $y^{\prime} + y f^{\prime}(x) - f(x) f^{\prime}(x) = 0$ and $\lim_{x \rightarrow \infty} y(x) = 0$, then (where $y^{\prime} = \frac{dy}{dx}$):

Let $y = y(x)$ be the solution of the differential equation $x\sqrt{1-x^2} dy + (y\sqrt{1-x^2} - x\cos^{-1}x) dx = 0$, where $x \in (0, 1)$ and $\lim_{x\to 1^-} y(x) = 1$. Then $y\left(\frac{1}{2}\right)$ equals:

The graph of the function $y = f(x)$ passing through the point $(0, 1)$ and satisfying the differential equation $\frac{dy}{dx} + y \cos x = \cos x$ is such that

If $y = A(x) e^{\int P dx}$ is a solution of $\frac{dy}{dx} + P(x) y = Q(x)$,then $A'(x) =$