The general solution of $\cos ^2 x \frac{d y}{d x}+y=\tan x$ is

Let $y^{\prime}(x) + y(x) g^{\prime}(x) = g(x) g^{\prime}(x)$,$y(0) = 0$,$x \in \mathbb{R}$,where $f^{\prime}(x)$ denotes $\frac{d f(x)}{d x}$ and $g(x)$ is a given non-constant differentiable function on $\mathbb{R}$ with $g(0) = g(2) = 0$. Then the value of $y(2)$ is

Solve the differential equation $\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{dy}=1$ where $x \neq 0$.

Let $y=y(x)$ be the solution of the differential equation $(2x \ln x) \frac{dy}{dx} + 2y = \frac{3}{x} \ln x$,$x > 0$ and $y(e^{-1}) = 0$. Then,$y(e)$ is equal to

Suppose $y=y(x)$ is the solution curve to the differential equation $\frac{dy}{dx}-y=2-e^{-x}$ such that $\lim_{x \rightarrow \infty} y(x)$ is finite. If $a$ and $b$ are respectively the $x$- and $y$-intercepts of the tangent to the curve at $x=0$,then the value of $a-4b$ is equal to:

The integrating factor of the differential equation $x \cdot \frac{dy}{dx} + 2y = x^2$ is $(x \neq 0)$.