Solve the equation $y' + (xy)^2 = -2/(x^4)$ knowing that $y_1 = 1/(x^3)$ is the particular solution

Question

$$y' + (xy)^2 = -\frac2{x^4}\text{ with }y_1 = \frac1{x^3}$$

I have tried using various differential equations methods to solve but it appears to be very challenging.

Answer 1

Your differential equation is a Riccati equation. The general form of this type is given by $$y'+p(x)y+q(x)y^2=r(x)$$ The attempt is to set $y=y_1(x)+v(x)$ where $y_1(x)$ is a particular solution. In your case choose $y=\frac1{x^3}+v(x)$ and plug this in your equation. Considering $p(x)=x^2,q(x)=1$ and $r(x)=-\frac2{x^2}$ this yields to an ODE in $v(x)$ an just evaluate $y$ afterwards by adding the particular solution.

But there is a problem in your case; your given $y_1$ does not fulfill the DE since one would get

$$\begin{align} \left(-\frac3{x^4}\right)+\left(\frac{x}{x^3}\right)^2=\frac1{x^4}-\frac3{x^4}=-\frac{2}{x^4}\ne-\frac2{x^2} \end{align}$$

Therefore $y_1=\frac1{x^3}$ would be a solution if and only if the RHS is given by $-\frac2{x^4}$.

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For clarification we are dealing with the following problem

$$y'+(xy)^2=-\frac2{x^4}\text{ with }y_1=\frac1{x^3}$$

Hence this DE is a Ricatti equation we make the attempt $y=y_1(x)+v(x)$ with the given particular solution. Plugging $y$ and $y'$ within the equation yields to

$$\begin{align} \left(\frac1{x^3}+v(x)\right)'+\left(\frac1{x^2}+xv(x)\right)^2&=-\frac2{x^4}\\ -\frac3{x^4}+v'(x)+\frac1{x^4}+x^2v^2(x)+2\frac{v(x)}x&=-\frac2{x^4}\\ v'(x)+2\frac{v(x)}x+x^2v^2(x)&=0 \end{align}$$

The last equation is now a Bernoulli equation therefore we divide by $-v^2(x)$ perform the substitution $z=\frac1{v(x)}$ and $z'=-\frac{v'(x)}{v^2(x)}$ to get

$$\begin{align} v'(x)+2\frac{v(x)}x+x^2v^2(x)&=0\\ \Leftrightarrow -\frac{v'(x)}{v^2(x)}-2\frac1{xv(x)}-x^2&=0\\ \Leftrightarrow z'-2\frac zx-x^2&=0\\ \end{align}$$

The last equation is a simple ODE in $z$ and can be solved by standard techniques. I will leave to you to verify that one will get

$$z=cx^2+x^3\Rightarrow v(x)=\frac1{cx^2+x^3}$$

and so we finally get

$$y=y_1+v(x)=\frac1{x^3}+\frac1{cx^2+x^3}$$

as the general solution.

Answer 2

Equations of the type $$ y'+(xy)^2=ax^b $$ have solutions of the form $y=cx^d$ if $$ cdx^{d-1}+c^2x^{2d+2}=ax^b $$ This requires all three exponents to be equal, $d-1=2d+2=b$, so that $d=-3$, $b=-4$. Then necessarily $a=c(c-3)$, which with the choice $c=1$ gives $a=-2$, as you finally gave in the correction of your equation.

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Sometimes increasing the order leads to an easier to solve problem, if there is a nice solution at all. So set $y=\frac{u'}{x^2u}$, then $$ -\frac2{x^4}=y'+(xy)^2=\frac{u''}{x^2u}-\frac{2u'}{x^3u}-\frac{u'^2}{x^2u^2}+\frac{u'^2}{x^2u^2} \implies x^2u''-2xu'+2u=0 $$ which indeed has easy solutions, for instance as Euler-Cauchy DE. You also know that $u_1=x$ is one solution, so that you can use order reduction.