The particular solution of $e^{\frac{y}{x}} = x, y(1) = 3, x > 0$ is . . . . . . .

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

Let $y=y(x), x>1$,be the solution of the differential equation $(x-1) \frac{d y}{d x}+2 x y=\frac{1}{x-1}$,with $y(2)=\frac{1+e^{4}}{2 e^{4}}$. If $y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}}$,then the value of $\alpha+\beta$ 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 $....$

The general solution of the differential equation $\frac{dy}{dx} = y \tan x - y^2 \sec x$ is

The solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ $(x \neq 0)$ with $y(1) = 1$ is