Question

If p = dy/dx, show that d^2y/ dx^2 = p(dp/dy).
Hence find the solution y = f(x) of the differential equation y(d^2y/dx^2) = 2(dy/dx) + (dy/dx)^2.

Answer 1

$$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(\dfrac{dy}{dx})=\dfrac{dp}{dx}=\dfrac{dp}{dy}(\dfrac{dy}{dx})=p\dfrac{dp}{dy}$$

$$p=\dfrac{dy}{dx}$$

Then

$$y(p\dfrac{dp}{dy})=2p+p^2$$

$$p=0\ or\ y\dfrac{dp}{dy}=p+2$$

$p=0=>\dfrac{dy}{dx}=0=>y=C_1$

$$\dfrac{dp}{p+2}=\dfrac{dy}{y}$$

$$\int\dfrac{dp}{p+2}=\int\dfrac{dy}{y}$$

$$p+2=C_2y$$

$$\dfrac{dy}{dx}=C_2y-2$$

$$\dfrac{dy}{C_2y-2}=dx$$

$$\dfrac{1}{C_2}\ln|C_2y-2|=x+\dfrac{1}{C_2}\ln C_3$$

$$C_2y-2=C_3e^{C_2x}$$

$$y=\dfrac{2}{C_2}+C_4e^{C_2x}, C_2\not=0$$

$$y=-2x+C_5, C_2=0$$

$y=C_1$

$y=-2x+C_5$

$y=\dfrac{2}{C_2}+C_4e^{C_2x}, C_2\not=0$