The general solution of the differential equation $\frac{d^2 y}{d x^2}+8 \frac{d y}{d x}+16 y=0$ is

The differential equation $\frac{dy}{dx} + \frac{1}{x} \sin 2y = x^3 \cos^2 y$ represents a family of curves given by:

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

Let $y(x)$ be a solution of $(1+x^{2}) \frac{dy}{dx} + 2xy - 4x^{2} = 0$ and $y(0) = -1$. Then $y(1)$ is equal to

Let $y=y(x)$ be the solution of the differential equation $\sec x \frac{dy}{dx} - 2y = 2 + 3 \sin x$, where $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$ and $y(0) = -\frac{7}{4}$. Then $y(\frac{\pi}{6})$ is equal to:

If the curve $y=y(x)$ is the solution of the differential equation $2(x^{2}+x^{5/4}) dy - y(x+x^{1/4}) dx = 2x^{9/4} dx, x > 0$ which passes through the point $(1, 1-\frac{4}{3} \log_{e} 2)$,then the value of $y(16)$ is equal to :