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My Problem is, this given second order non-linear differential equation: $$yy^{\prime\prime}=2(y^{\prime})^{2}-2y^{\prime}$$

I am really stuck with second-order differential equations.

My Approach: i was given the advise to solve first $$y^{\prime}=p(y)$$ and to determine $p$ but i didn't managed to get it.

I thought this: $$yy^{\prime\prime}=2(y^{\prime})^{2}-2y^{\prime}$$ could lead to: $$2y^{\prime}=2(y^{\prime})^{2}-yy^{\prime\prime}$$ $$y^{\prime}=(y^{\prime})^{2}-\frac{yy^{\prime\prime}}{2}$$

But i don't get it and i can't solve the equation.

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  • $\begingroup$ @TonyPiccolo: I think you mean $\dfrac{\tan\left(\sqrt{C_1} (x+C_2)\right)}{\sqrt{C_1}}$ $\endgroup$ Jun 24, 2013 at 11:33
  • $\begingroup$ $y''=\frac{dy'}{dx} = \frac{dy}{dx}\frac{dy'}{dy}$ $\endgroup$
    – Empy2
    Jun 24, 2013 at 11:35
  • $\begingroup$ @Robert Israel: Yes, thank you. I removed the comment because I was not sure. $\endgroup$ Jun 24, 2013 at 16:11

2 Answers 2

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Hint: You can write your differential equation in the form $$ \dfrac{y''}{y' - 1} = 2 \dfrac{y'}{y} $$ and both sides can be integrated...

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  • $\begingroup$ No, $\int 2 \dfrac{y'}{y}\ dx = 2 \ln y + C$. And similarly for $\int \dfrac{y''}{y'-1}\ dx$. $\endgroup$ Jun 24, 2013 at 14:47
  • $\begingroup$ oh okay. But $\int \dfrac{y''}{y'-1}\ dx=ln(y'-1)+C$... and this would result in $$ln(y'-1)+C=2 \ln y+C$$ ... right? $\endgroup$ Jun 24, 2013 at 19:25
  • $\begingroup$ Right (of course, the two $C$'s are not necessarily equal). Now exponentiate to get rid of the logs. $\endgroup$ Jun 24, 2013 at 20:02
  • $\begingroup$ okay, should i get this in the end? $$y=\tan (x+C)$$ $\endgroup$ Jun 24, 2013 at 20:11
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    $\begingroup$ $y' = 1 + A y^2$ is a separable differential equation. $\endgroup$ Jun 26, 2013 at 6:46
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ODE has trivial solution $y=0$.

Now we have to solve case $y\neq 0$:

After dividing equation with $y^3$: $$\frac{y y''-2y'^2}{y^3}=\frac{-2y'}{y^3},$$ we get $(\frac{y'}{y^2})'$ on the left side and $(\frac{1}{y^2})'$ on the right side. Therefore, $$\frac{y'}{y^2}=\frac{1}{y^2}+C,$$ $$y'=1+Cy^2,$$ $$\frac{dy}{1+Cy^2}=dx.$$ After integration, we have $$\frac{1}{\sqrt{C}}\tan^{-1}(\sqrt{C}y)=x+D.$$ Finally, $$\frac{1}{\sqrt{C}}\tan{(\sqrt{C}(x+D))}=y.$$

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