The solution of $\frac{d^2y}{dx^2} = \cos x - \sin x$ 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

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

Let $y$ be the solution of the differential equation $x \frac{dy}{dx} = \frac{y^2}{1 - y \log x}$ satisfying $y(1) = 1$. Then, $y$ satisfies:

If $f(x), f^{\prime}(x), f^{\prime \prime}(x)$ are positive functions and $f(0)=1, f^{\prime}(0)=2$,then the solution of the differential equation $\left|\begin{array}{ll}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}\right|=0$ is

$A$ solution curve of the differential equation $(x^2+xy+4x+2y+4) \frac{dy}{dx}-y^2=0, x>0$,passes through the point $(1,3)$. Then the solution curve