The particular solution of the differential equation $xdy = (2x^2 + 1)dx, x \neq 0$, given that $y = 1$ when $x = 1$ is:

The general solution of the given differential equation $\frac{dy}{dx} = \frac{3e^{2x}+3e^{4x}}{e^{x}+e^{-x}}$ (where C is an arbitrary constant)

The solution of the differential equation $\frac{dr}{dt} = -rt, r(0) = r_0$ is

The solution of the differential equation $(x + 1)\frac{dy}{dx} + 1 - 2e^{-y} = 0$, $y(0) = 0$ is

The solution of differential equation $\sqrt{x+1} - \sqrt{x-1}\frac{dy}{dx} = 0$ is

The product of the order and degree of the differential equation $\left(\frac{d^3y}{dx^3}\right)^2 - 3\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^4 = y^4$ is

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