The solution of $y\,dx - x\,dy + 3x^2y^2e^{x^3}dx = 0$ is

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

If $y$ is a function of $x$,then $\frac{d^2y}{dx^2} + y \frac{dy}{dx} = 0$. If $x$ is a function of $y$,then the equation becomes:

If the transformation $z = \log \tan \frac{x}{2}$ reduces the differential equation $\frac{d^2 y}{d x^2} + \cot x \frac{d y}{d x} + 4 y \operatorname{cosec}^2 x = 0$ into the form $\frac{d^2 y}{d z^2} + k y = 0$, then $k$ is equal to

If $2xy^3dx + x^2y^2dy = ydx - xdy$ and $y(2) = 1$,then the value of $y(-1)$ will be (where $y(x)$ denotes the value of $y$ for a given $x$):

Verify that the given function $y - \cos y = x$ is a solution of the differential equation $(y \sin y + \cos y + x) y' = y$.