Solving a non linear differential equation using Perturbation

Question

$$yy'+x+\lambda \frac{x}{y}=0$$ Where $\lambda$ is very small and $y(0)=R$.

How can we solve it with Substituting: $y=y_0+\lambda y_1 +\lambda^2 y_2+\cdots$

Also, can we solve it without this method?

Answer 1

There is an implicit solution

$$ {x}^{2}+ y \left( x \right) ^{2}-2\,y \left( x \right) \lambda+2\,{\lambda}^{2}\ln \left( \lambda+y \left( x \right) \right) = {R}^{2}-2\,R\lambda+2\,{\lambda}^{2}\ln \left( \lambda+R \right) $$

Now for the series solution.

If we substitute $y = y_0 + \lambda y_1 + \lambda^2 y_2 + \ldots$ into the differential equation, we get

$$ y_0 y_0' + x + \left(y_0 y_1' + y_0' y_1 + \dfrac{x}{y_0}\right) \lambda + \left(y_0 y_2' + y_1 y_1' + y_0' y_2 - \dfrac{x y_1}{y_0^2}\right) \lambda^2 + \ldots = 0$$

The $\lambda^0$ term is $y_0 y_0' + x = 0$, whose solution (with $y_0(0) = R$) is $y_0 = \sqrt{R^2 - x^2}$ if $R > 0$, or $-\sqrt{R^2 - x^2}$ if $R < 0$.

Each subsequent term gives us a first-order linear differential equation for $y_i$ in terms of $y_j$ for $j < i$, which we solve with the initial condition $y_i(0) = 0$. Thus for $i=1$ the solution of $y_0 y_1' + y_0' y_1 + x/y_0 = 0$ with $y_1(0)=0$ (in the case $R > 0$) is $y_1 = 1 - \dfrac{R}{\sqrt{R^2 - x^2}}$.

Answer 2

Exact solution (without series expension) : $$yy'+x+\lambda\frac{x}{y}=0$$ $$x=-\frac{y^2}{y+\lambda}y'$$ $$\int x\:dx=-\int\frac{y^2}{y+\lambda}y'$$ $$\frac{x^2}{2}=\lambda y -\frac{y^2}{2} -\lambda^2 \ln|y+\lambda|+C$$ $y(0)=R\quad\to\quad C=-\lambda R +\frac{R^2}{2} +\lambda^2 \ln|R+\lambda|$ $$\frac{x^2}{2}=\lambda (y-R) -\frac{y^2-R^2}{2} -\lambda^2 \ln|\frac{y+\lambda}{R+\lambda}|$$ If you want only the two first terms of the serie expension relatively to $\lambda$ it is simpler to derive it directly from the ODE : Replace $y(x)$ by $y_0(x)+y_1(x)\lambda+y_2(x)\lambda^2 $ into the ODE.

It is more complicated from the exact solution above : expend the Relationship in terms of powers of $\lambda$ and you will obtain : $$y_0=\pm\sqrt{R^2-x^2}$$ $$y_1=1-\frac{R}{y_0}$$ $$y_2=\frac{y_1-\frac{y_1^2}{2}-\ln|\frac{y_0}{R}|}{y_0}$$