Question

Solve y+px=p²x⁴

Answer 1

The given differential equation is

$$y+px=p^2x^4$$

Differentiating with respect to $x,$ we get 

$$\dfrac{dy}{dx}+p+x\dfrac{dp}{dx}=2px^4\dfrac{dp}{dx}+4p^2x^3$$

$$p+p+x\dfrac{dp}{dx}=2px^4\dfrac{dp}{dx}+4p^2x^3$$

$$\dfrac{dp}{dx}(x-2px^4)=4p^2x^3-2p$$

$$x(1-2px^3)\dfrac{dp}{dx}=-2p(1-2px^3)$$

$$1-2px^3=0$$

$$p=\dfrac{1}{2x^3}$$

Substitute

$$y+(\dfrac{1}{2x^3})x=(\dfrac{1}{2x^3})^2x^4$$

$$y=-\dfrac{1}{2x^2}+\dfrac{1}{4x^2}$$

$$y=-\dfrac{1}{4x^2}$$

$$x\dfrac{dp}{dx}=-2p$$

$$\dfrac{dp}{p}+2\dfrac{dx}{x}=0$$

Integrating, we get

$$\ln|p|+\ln x^2=\ln C$$

$$px^2=C$$

Substitute

$$px=\dfrac{C}{x}$$

$$(px^2)^2=p^2x^4=C^2$$

$$y+px=p^2x^4$$

Then

$$y+\dfrac{C}{x}=C^2$$