The solution of $(x+y+1) \frac{dy}{dx} = 1$ is

If $x dy + (y + y^2 x) dx = 0$ and $y = 1$ at $x = 1$,then

If $y=y(x)$ is the solution of the equation $e^{\sin y} \cos y \frac{dy}{dx} + e^{\sin y} \cos x = \cos x$ with $y(0)=0$,then $1 + y\left(\frac{\pi}{6}\right) + \frac{\sqrt{3}}{2} y\left(\frac{\pi}{3}\right) + \frac{1}{\sqrt{2}} y\left(\frac{\pi}{4}\right)$ is equal to

If $\cos x$ and $\sin x$ are solutions of the differential equation $a_{0} \frac{d^{2} y}{d x^{2}}+a_{1} \frac{d y}{d x}+a_{2} y=0$ where $a_{0}, a_{1}$ and $a_{2}$ are real constants, then which of the following is/are always true?

The general solution of the differential equation $\frac{dy}{dx} + \frac{y \ln y}{x} = \frac{y(\ln y)^2}{x^2}$ is (where $C$ is an arbitrary constant):

Let $f$ be a real-valued differentiable function on $\mathbb{R}$ (the set of all real numbers) such that $f(1)=1$. If the $y$-intercept of the tangent at any point $P(x, y)$ on the curve $y=f(x)$ is equal to the cube of the abscissa of $P$,then the value of $f(-3)$ is equal to