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

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 3(\tan^2 x + 1)y = \sec^2 x$,with the initial condition $y(0) = \frac{1}{3} + e^3$. Then $y\left(\frac{\pi}{4}\right)$ is equal to:

Let the solution curve $y=f(x)$ of the differential equation $\frac{dy}{dx}+\frac{xy}{x^{2}-1}=\frac{x^{4}+2x}{\sqrt{1-x^{2}}}, x \in(-1,1)$ pass through the origin. Then $\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$ is equal to

Let $y=y(x)$ be the solution of the differential equation $x\frac{dy}{dx}-\sin(2y)=x^{3}(2-x^{3})\cos^{2}y,$ for $x\ne0.$ If $y(2)=0,$ then $\tan(y(1))$ 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 $y=f(x)$ is the solution of the differential equation $(1+\cos^2 x) f'(x) - f(x) \sin 2x = 4 \sin 2x$ with $f(0)=0$, then $f(\frac{\pi}{3})=$