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:

$A$ function $y = f(x)$ satisfies the differential equation $\frac{dy}{dx} - y = \cos x - \sin x$ with the initial condition that $y$ is bounded when $x \rightarrow \infty$. The area enclosed by $y = f(x)$,$y = \cos x$,and the $y$-axis is

Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t-x) dt - \int_{0}^{x} f(t) \tan t dt$,where $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then $f''\left(\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to . . . . . . .

The solution of the differential equation $x dy = (y + xy^3 (1 + \log_e x)) dx$ is (where $C$ is an arbitrary constant):

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 $\frac{d^2y}{dx^2} + \sin x = 0$,then the solution of the differential equation is ...... .