Let $y$ be the solution of the differential equation $(1-x^{2}) dy = (xy + (x^{3}+2) \sqrt{1-x^{2}}) dx$ for $-1 < x < 1$ with $y(0)=0$. If $\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1-x^{2}} y(x) dx = k$,then $k^{-1}$ is equal to:

For all $x > 0$,let $y_1(x), y_2(x)$,and $y_3(x)$ be the functions satisfying $\frac{dy_1}{dx} - (\sin x)^2 y_1 = 0, y_1(1) = 5$; $\frac{dy_2}{dx} - (\cos x)^2 y_2 = 0, y_2(1) = \frac{1}{3}$; and $\frac{dy_3}{dx} - \left(\frac{2-x^3}{x^3}\right) y_3 = 0, y_3(1) = \frac{3}{5e}$ respectively. Then $\lim_{x \rightarrow 0^{+}} \frac{y_1(x) y_2(x) y_3(x) + 2x}{e^{3x} \sin x}$ is equal to:

Suppose $y=y(x)$ is the solution curve to the differential equation $\frac{dy}{dx}-y=2-e^{-x}$ such that $\lim_{x \rightarrow \infty} y(x)$ is finite. If $a$ and $b$ are respectively the $x$- and $y$-intercepts of the tangent to the curve at $x=0$,then the value of $a-4b$ is equal to:

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

Let $x = x(y)$ be the solution of the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$. If $x(1) = 1$,then $x(\frac{1}{2})$ is:

If $y=y(x)$ is the solution of the differential equation $e^{y}\left(\frac{dy}{dx}-1\right)=e^{x}$ such that $y(0)=0,$ then $y(1)$ is equal to