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

Solve the differential equation $\left( {1 + x\sqrt {{x^2} + {y^2}} } \right)\,dx + \left( {\sqrt {{x^2} + {y^2}} - 1 } \right)y\,dy = 0$.

Suppose $f(x) = e^{ax} + e^{bx},$ where $a \neq b,$ and that $f''(x) - 2f'(x) - 15f(x) = 0$ for all $x.$ Then the product $ab$ is equal to

If the transformation $z = \log \tan \frac{x}{2}$ reduces the differential equation $\frac{d^2 y}{d x^2} + \cot x \frac{d y}{d x} + 4 y \operatorname{cosec}^2 x = 0$ into the form $\frac{d^2 y}{d z^2} + k y = 0$, then $k$ is equal to

The solution of the differential equation $y dx - x dy + 3x^2 y^2 e^{x^3} dx = 0$ satisfying $y = 1$ when $x = 1$ is:

Every curve represented by the general solution of $\frac{dy}{dx} = \frac{x \log x}{y^3 e^{y^2-5}}$ cuts every curve represented by the general solution of $\frac{dy}{dx} + \frac{y^3 e^{y^2-5}}{x \log x} = 0$ at an angle $\theta$. Then,$4\theta - \frac{\pi}{2} =$