The general solution of the differential equation $\frac{dy}{dx} + \left(\frac{3x^2}{1+x^3}\right)y = \frac{1}{x^3+1}$ is

Let $y=y(x)$ be the solution curve of the differential equation $\sin(2x^2) \ln(\tan x^2) dy + (4xy - 4\sqrt{2}x \sin(x^2 - \frac{\pi}{4})) dx = 0$ for $0 < x < \sqrt{\frac{\pi}{2}}$,which passes through the point $(\sqrt{\frac{\pi}{6}}, 1)$. Then $|y(\sqrt{\frac{\pi}{3}})|$ is equal to $.....$

The solution of the differential equation,$x^2 \frac{dy}{dx} \cos \frac{1}{x} - y \sin \frac{1}{x} = -1,$ where $y \rightarrow -1$ as $x \rightarrow \infty$ is

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:

Let $x = x(y)$ be the solution of the differential equation $2y^2 \frac{dx}{dy} - 2xy + x^2 = 0, y > 1, x(e) = e$. Then $x(e^2)$ is equal to:

Let $y=y(x)$ be the solution of the differential equation $(x-x^{3}) dy=(y+yx^{2}-3x^{4}) dx, x>2$. If $y(3)=3$,then $y(4)$ is equal to :