Non-Regular Differential Equation soultion

Question

I am currently a freshman at Japanese University and studying ordinary differential equation. I faced one problem which I could not solve.

$$y=-xp+x^4p^2$$ $$p=\frac{dy}{dx}$$

Can someone teach me how to solve this equation?

Answer 1

It's a bit of a trick, but the way to solve this is to interchange the dependent and independent variable. So, consider the differential equation as an equation for $x(y)$, using $\frac{\text{d} y}{\text{d} x} = \left(\frac{\text{d} x}{\text{d} y}\right)^{-1}$. Then, using the substitution $$ \xi(y) = 4 y^2 x(y) -1, $$ you'll have no problem solving the differential equation.

Answer 2

Another way to solve problems like this one.

From the second equation, changing the variable to $p$:

$$\frac{dy}{dx}=\frac{dy}{dp}\frac{dp}{dx}=p$$

From the first equation we find the first derivative:

$$\frac{dy}{dp}=-p\frac{dx}{dp} -x+2p x^4+4p^2 x^3 \frac{dx}{dp}$$

Substituting:

$$\frac{dy}{dp}\frac{dp}{dx}=-p-x\frac{dp}{dx}+2x^4 p \frac{dp}{dx}+4x^3 p^2=(4x^3p-1)p+x(2x^3p-1)\frac{dp}{dx}$$

From the second equation again:

$$(4x^3p-1)p+x(2x^3p-1)\frac{dp}{dx}=p$$

$$2(2x^3p-1)p+x(2x^3p-1)\frac{dp}{dx}=0$$

$$(2x^3p-1) \left(2p+x \frac{dp}{dx} \right)=0$$

This separates into two equations, one algebraic and one separable ODE.

Answer 3

Set $x=u^{-1}$, $v(u)=y(x)=y(u^{-1})$, then $y'(x)=-x^{-2}v'(u)=-u^2v'(u)$ and in consequence $$ v(u)=uv'(u)+v'(u)^2 $$ which now is a classical Clairaut differential equation with derivative $0=v''(u+2v')$. It has linear solutions $$ v(u)=cu+c^2\iff y(x)=\frac{c}x+c^2 $$ and a singular solution given by $u+2v'(u)=0$, that is $$ v(u)=-\frac{u^2}4\iff y(x)=-\frac1{4x^2}. $$ As the first family is tangent to the singular solution, one can also assemble piecewise solutions that move for some time along the singular solution and continue before and/or after along a solution of the family.