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Thequestion is to find out a solution of differential equation $$x^2+y^2+2(\frac{dy}{dx})^2+2y\frac{d^2y}{dx^2}+2=0$$

I noticed that $\frac{d^2(x^2+y^2)}{dx^2}=2(\frac{dy}{dx})^2+2y\frac{d^2y}{dx^2}+2$So that the differential equation reduces to $$x^2+y^2+\frac{d^2(x^2+y^2)}{dx^2}=0$$I couldn't proceed after this.Any ideas.thanks.

catechol
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2 Answers2

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let $v = x^2 + y^2$ $$v + \frac{d^2v}{dx^2} = 0$$

$$v = c_1\sin x + c_2 \cos x$$ $$ y^2 = c_1\sin x + c_2 \cos x -x^2$$

EDIT:

As requested by OP Ill show why the solution to $v ''+ v =0$

We first start by assuming $v = e^{kx}$ for some unknown constant $k$. We substitute this back into our differential equation to get $$k^2e^{kx} + e^{kx} = 0$$ Which is true only when $k = \pm i$ so the solution can be written as $$ v= c_1e^{-ix} + c_2e^{ix}$$ However it is generally more appropriate to avoid complex numbers so we find another basis for our solution

One being when $c_1 = c_2 = \frac{1}{2}$ which gives us $\frac{e^{ix} + e^{-ix}}{2} = \cos x$

Another being when $c_1 = 1/2$, $c_2 = -1/2i$ which gives us $\frac{e^{ix} - e^{-ix}}{2i} = \sin x$ So this therefore gives us our new basis.

Ziad Fakhoury
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  • Is that a standard differential equation?how do one prove that?thanks. – catechol Apr 06 '17 at 16:12
  • I edited to show you how it is standardly shown, however you can also very simply notice that for $\sin$ and $\cos$ their second derivative is the negative of them. – Ziad Fakhoury Apr 06 '17 at 16:23
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Hint: So if $u = x^2 + y^2$, you need to solve $u'' + u = 0$. Surely you know how to do that?

Robert Israel
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