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We want to solve the differential equation :

  • $yy''-(y')^2=-y^2$
  • $y(0)=1, y'(0)=0$

  • $ \left(\frac{y'}{y}\right)'=\frac{yy''-y'^2}{y^2}=-1$
  • therefore $y(x) =e^{ \frac{-x^2}2}$

How would you solve it, without trick, with a general method ?

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2 Answers 2

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Since the ODE is of the form: $$\Psi(y,y',y'')=0\ $$ You can substitute $p=\dfrac {dy}{dx}$ and $y''=p\dfrac {dp}{dy} $. So that: $$yy''-(y')^2=-y^2$$ Becomes: $$yp'-p=-\dfrac {y^2}{p}$$ This is Bernoulli's differential equation.

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A general second order ODE has the form $$\Psi(x,y,y',y'')=0\ ,$$ and your problem is of this sort. There are some special such equations, e.g., linear ODEs with constant coefficients, for which there is a standard algorithm. But there is no general method that solves any ODE of the above form. When you find a solution nevertheless this is not a trick, but luck.

Concerning the ODE at hand you had exactly the relevant special idea that solves it.

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  • $\begingroup$ Thanks so much ! $\endgroup$
    – zestiria
    Apr 23, 2021 at 12:56

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