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I was reading this Wikipedia article. I tried to adopt the same method that they used to solve the following diffrential equation:

$$y\frac {d^2(y)}{dx^2}+{(\frac {dy}{dx})}^2+y^2=0$$ Putting $y=e^{rx}$ and then taking out $e^{2rx}$ outside with $2r^2+1$ inside. Solving for $r$ we get $y=e^{\frac{ix}{\sqrt 2}}$ or $y=e^{-\frac{ix}{\sqrt 2}}$ and when we combine both results as one and apply euler form of complex number we get $$y=a\cos {\frac{x}{\sqrt 2}} + b\sin {\frac{x}{\sqrt 2}}$$ but when I used Wolfram Alpha (computational software) to verify my solution by plugging in the solution into the diffrential equation I don't get $0$ but $1$. I am not able to identify what I did wrong. It would be very nice if you could help me with identifying my mistake. Thanks a lot.

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    $\begingroup$ Hint: Instead, try $$ p = y' \implies p' = y''$$ Substitute those two and solve and then substitute again. $\endgroup$ – Moo Oct 15 '18 at 19:22
  • $\begingroup$ your equation is not linear $\endgroup$ – LostInSpace Oct 15 '18 at 19:54
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Your equation is not linear...

$$y\frac {d^2(y)}{dx^2}+{(\frac {dy}{dx})}^2+y^2=0$$ $$\color {red} {y''y+(y')^2}+y^2=0$$ $$ \color {red}{(y'y)'}+y^2=0$$ $$ (2y'y)'+2y^2=0$$ $$((y^2))''+2y^2=0$$ Substitute $u=y^2$ $$u''+2u=0$$ This is linear.... $$r^2+2=0 \implies r=\pm i\sqrt 2$$ $$\implies u=c_1\cos(\sqrt 2 x)+c_2 \sin ( \sqrt 2 x)$$ $$\boxed {y^2(x)=c_1\cos(\sqrt 2 x)+c_2 \sin ( \sqrt 2 x)}$$


Another method is

$$y'=\frac {dy}{dx}=p $$ And $$y''=\frac {dp}{dx}=\frac {dp}{dy}\frac {dy}{dx}=pp'_y$$ The equation becomes $$pp'y+p^2+y^2=0$$ Note that $(p^2)'=2pp'$ $$\frac 12 (p^2)'y+(p^2)+y^2=0$$ Substitute $u=p^2$ $$ u'+2\frac uy=-2y$$ Its linear of first order...

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  • $\begingroup$ I solved the linear first order. I get the same answer as I got by wikipedia method. I think wolfram alpha is giving wrong answer. $\endgroup$ – Mayank Mittal Oct 15 '18 at 20:25
  • $\begingroup$ Wa gives the right answer...you sure you put the right equation ? @infinitelycurious $\endgroup$ – LostInSpace Oct 15 '18 at 20:26
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    $\begingroup$ @infinitelycurious I don't know but type just $y'$ for $ \frac {dy}{dx} $ I typed the equation in wa wolframalpha.com/input/?i=yy%27%27%2B(y%27)%5E2%2By%5E2%3D0 $\endgroup$ – LostInSpace Oct 15 '18 at 20:31
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    $\begingroup$ I think that del stuff which it is accepting instead of dy/dx is making the diff. Now I feel bad for wasting your time as well as mine but atleast I learned a few new methods to solve DE. $\endgroup$ – Mayank Mittal Oct 15 '18 at 20:38
  • $\begingroup$ @infinitelycurious no problem we are here here to help one another...so it was my pleasure to answer your question...Take care $\endgroup$ – LostInSpace Oct 15 '18 at 20:40
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Let $$v(y)=\frac{dy(x)}{dx}$$ then $$2\frac{dv(y)}{dy}v(y)+\frac{2v(y)^2}{y}=-2y$$ Now let $$u(y)=v(y)^2$$ then $$\frac{du(y)}{dy}+\frac{u(y)}{y}=-2y$$ Can you proceed?

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  • $\begingroup$ You did a calculation mistake. It must be 2u(y)/y. Although, I get the same answer as what I got before through that wikipedia method. Is there a possibility that Wolfram aplha is giving wrong answer? $\endgroup$ – Mayank Mittal Oct 15 '18 at 20:23
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If you look sharp enough, the first two terms have the form of a product derivative. Indeed $(yy')'=yy''+(y')^2$. This in turn is the derivative of the square, so that in the end the ODE is equivalent to $$ (y^2)''+2(y^2)=0 $$ which is an oscillation equation for $u=y^2$ with solutions $$y^2=u=A\sin(\sqrt2 x+\phi).$$ As these sinoids also take negative values, the solution for $y$ has some sudden stops.

Or in initial conditions: $$ y(x)^2=y(0)^2\cos(\sqrt2 x)+\sqrt2y(0)y'(0)\sin(\sqrt2 x) $$ so that the square root with the correct sign is $$ y(x)=y(0)\sqrt{\cos(\sqrt2 x)+\sqrt2\frac{y'(0)}{y(0)}\sin(\sqrt2 x)}. $$

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  • $\begingroup$ your idea seems correct but if we follow isham's another method, we get a continious expression for y. Isnt that contradictory? I think A will turn negative in that case because if we conpare both of them, A is actually related to c1 and c2 and ince c1 and c2 are complex. A might take a negative value. $\endgroup$ – Mayank Mittal Oct 15 '18 at 20:43
  • $\begingroup$ If you allow complex values for $y$, then you need the more general solution formula. Our expressions are the same, that is equivalent forms, if $y$ is restricted to be a real function. $\endgroup$ – Lutz Lehmann Oct 15 '18 at 20:51

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