$A$ solution of $y = 2x\left( \frac{dy}{dx} \right) + x^2\left( \frac{dy}{dx} \right)^4$ is

Let the tangent at any point $P(x, y)$ on a curve passing through the points $(1, 1)$ and $(\frac{1}{10}, 100)$ intersect the positive $x$-axis and $y$-axis at the points $A$ and $B$ respectively. If $PA: PB = 1: k$ and $y = y(x)$ is the solution of the differential equation $e^{\frac{dy}{dx}} = 2x + 1$ with $y(0) = 2$,then $4y(1) - 5 \log_e 3$ is equal to:

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

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 . . . . . . .

$A$ solution of the differential equation ${\left( {\frac{{dy}}{{dx}}} \right)^2} - x\frac{{dy}}{{dx}} + y = 0$ is

The solution of $(1 + xy)y\,dx + (1 - xy)x\,dy = 0$ is