The solution of the differential equation $e^{-x}(y+1) dy + (\cos^2 x - \sin 2x) y dx = 0$ at $x=0, y=1$ is

The solution of $y{e^{ - x/y}}dx - (x{e^{ - x/y}} + {y^3})dy = 0$ is

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

$A$ continuously differentiable function $\phi (x)$ in $(0, \pi)$ satisfying $y' = 1 + y^2$ and $y(0) = 0 = y(\pi)$ is

Consider the family of all circles whose centers lie on the straight line $y = x$. If this family of circles is represented by the differential equation $P y^{\prime \prime} + Q y^{\prime} + 1 = 0$,where $P, Q$ are functions of $x, y$ and $y^{\prime}$ (here $y^{\prime} = \frac{dy}{dx}, y^{\prime \prime} = \frac{d^2y}{dx^2}$),then which of the following statements is (are) true?
$(A) P = y + x$
$(B) P = y - x$
$(C) P + Q = 1 - x + y + y^{\prime} + (y^{\prime})^2$
$(D) P - Q = x + y - y^{\prime} - (y^{\prime})^2$

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