Second order non linear ODE - hard to solve integral makes me think I need a different substitution

Question

I have this here ODE:

$$xy'' = y' + x((y')^2 + x^2)$$

Naturally, I'd try this substitution first: $$y' = p, p=p(x)$$

The equation then transforms into $$xp' = p+x(p^2+x^2)$$ Dividing it by $x$, I get

$$p' = \frac{p}{x} + p^2 + x^2$$

Which is a Riccati equation with the solution: $$p = x \cdot \tan{(\frac{x^2}{2}+C_1)}$$

The thing is, if I substitute back $y'=p$, the integral on the right side is not an easy one to solve, and even if I do solve it with WolframAlpha the solutions are not the same as if I plug in the second-order equation directly. It makes me wonder if I should have tried another substitution/method.

Any help will be appreciated!

Answer 1

$$y' = x \cdot \tan{\left(\frac{x^2}{2}+C\right)}$$ $$\dfrac {dy}{du}\dfrac {du}{dx} = x \cdot \tan(u)$$ $$\dfrac {dy}{du} = \tan(u)$$ Where $u=\dfrac {x^2}{2}+C$ then integrate. $$y=-\ln |\cos u |+K$$

Answer 2

You have now a formula of the form $$ y'(x)=f(u(x))u'(x) $$ with $f(u)=\tan(u)$ and $u(x)=\frac{x^2}2+c$. This the gives that $$ y(x)=F(u(x))+d, $$ where $F'=f$, here $F(u)=-\ln|\cos(u)|$

Answer 3

Let $y'=xz$. The equation becomes

$$xz+x^2z'=xz+x(x^2z^2+x^2),$$ which is separable:

$$\frac{z'}{z^2+1}=x$$ or $$\arctan z=\frac{x^2}2+c$$ and $$y'=x\tan\left(\frac{x^2}2+c\right)$$ as you found.

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Now the integral is not difficult: $$\int x\tan\left(\frac{x^2}2+c\right)dx=\int \tan\left(u+c\right)du=-\log(\cos(u+c))+c'\\ =-\log\left(\cos\left(\frac{x^2}2+c\right)\right)+c'.$$