The differential equation $y^2 dx + (2xy - 1) dy = 0$ is

Let $f$ be a differentiable function $f : R \rightarrow R$ satisfying the equation $f(x) = (1+x^2) \left[ 1 + \int_{0}^{x} \frac{f(t)}{1+t^2} dt \right]$ for all $x \in R$. Then $f(1)$ is:

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 2y = f(x)$,where $f(x) = \begin{cases} 1, & x \in [0, 1] \\ 0, & \text{otherwise} \end{cases}$. If $y(0) = 0$,then $y\left(\frac{3}{2}\right)$ is

Find the solution of the differential equation $(e^{y-x}) dy = (e^x - e^y) dx$.

The general solution of the differential equation $(1-x^{2}) \frac{dy}{dx} + 2xy = x(1-x^{2})^{\frac{1}{2}}$ is

If the length of the sub-tangent at any point $P(x, y)$ on a curve $f(x, y) = 0$ is $x + 7y^2$, then $f(x, y) =$