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

Let $y=y(x)$ be the solution of the differential equation $(x^2+1) y^{\prime}-2 x y=(x^4+2 x^2+1) \cos x$,with $y(0)=1$. Then $\int_{-3}^3 y(x) d x$ is :

Let $y = y_1(x)$ and $y = y_2(x)$ be the solution curves of the differential equation $\frac{dy}{dx} = y + 7$ with initial conditions $y_1(0) = 0$ and $y_2(0) = 1$ respectively. Then the curves $y = y_1(x)$ and $y = y_2(x)$ intersect at

Find the general solution of the differential equation: $\cos ^{2} x \frac{d y}{d x}+y=\tan x$ where $0 \leq x < \frac{\pi}{2}$.

$A$ function $y = f(x)$ satisfying the differential equation $\frac{dy}{dx} \sin x - y \cos x + \frac{\sin^2 x}{x^2} = 0$ is such that $y \rightarrow 0$ as $x \rightarrow \infty$. Which of the following statements is correct?

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