By multiplying with $e^{\int P dx}$ on both sides of the equation $\frac{dy}{dx} + P(x)y = Q(x)$,the left side of the equation takes the form $\frac{d}{dx}(y f(x))$,then $f(x) =$

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

If $x=x(t)$ is the solution of the differential equation $(t+1) dx = (2x + (t+1)^4) dt$ with the initial condition $x(0) = 2$,then $x(1)$ equals:

If $y=y(x)$ is the solution of the differential equation $\frac{dy}{dx}+\frac{4x}{x^2-1}y=\frac{x+2}{(x^2-1)^{5/2}}$ for $x > 1$,such that $y(2)=\frac{2}{9}\log_e(2+\sqrt{3})$ and $y(\sqrt{2})=\alpha\log_e(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}$,where $\alpha, \beta, \gamma \in N$,then $\alpha\beta\gamma$ is equal to $........$.

If the curve $y=y(x)$ is the solution of the differential equation $2(x^{2}+x^{5/4}) dy - y(x+x^{1/4}) dx = 2x^{9/4} dx, x > 0$ which passes through the point $(1, 1-\frac{4}{3} \log_{e} 2)$,then the value of $y(16)$ is equal to :

If the solution curve $f(x, y)=0$ of the differential equation $(1+\log_e x) \frac{dx}{dy} - x \log_e x = e^y, x > 0$,passes through the points $(1,0)$ and $(\alpha, 2)$,then $\alpha^\alpha$ is equal to