Innocent looking but confusing DE!

**usagi_killer** · December 15th 2010, 04:23 AM

Hmm am I missing something obvious but how would I solve y"=ay^2?

**Ackbeet** · December 15th 2010, 04:32 AM

Multiply both sides by y' and integrate. That'll reduce the order by one. What does that give you?

**chisigma** · December 15th 2010, 08:29 AM

Originally Posted by usagi_killer

Hmm am I missing something obvious but how would I solve y"=ay^2?

For this type of second order ODE the 'standard approach' is based on the identity...

$\displaystyle y^{''} = \frac{d y^{'}}{dx}= \frac{d y^{'}}{dy}\ \frac{dy}{dx} = y^{'}\ \frac{d y^{'}}{dy}$ (1)

... so that the equation you have prosed $y^{''}= a\ y^{2}$ becomes...

$\displaystyle y^{'}\ dy^{'} = a\ y^{2}\ dy$ (2)

... that is 'separated variables' and the solution of which is...

$\displaystyle y^{'\ 2}= \frac{2\ a}{3} y^{3} + c_{1} \implies y^{'} = \pm \sqrt{\frac{2\ a}{3}\ y^{3} + c_{1}}$ (3)

Also (3) is 'sepatated variables' and its solution is...

$\displaystyle x = \int \frac{d y}{\pm \sqrt{\frac{2\ a}{3}\ y^{3} + c_{1}}} + c_{2}$ (4)

... even if the integral in (4) is not very 'confortable'...

Merry Christmas from Italy

$\chi$ $\sigma$

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