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Differential Equations Ordinary and Partial Differential Equations Math Forum


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January 23rd, 2018, 03:42 AM   #1
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2*y''=3*y^2

I have some ideas about this equation, but nevertheless I'm stuck; please help.

Last edited by skipjack; January 23rd, 2018 at 01:02 PM.
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January 23rd, 2018, 01:09 PM   #2
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Use W|A.
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January 23rd, 2018, 03:54 PM   #3
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Try multiplying both sides by $y'$. My solution is
$$c^2-\frac{4c}{\sqrt y}+\frac4y=x^2$$
A different rearrangement of this is
$$y=\frac4{(c \pm x)^2}$$
but that may include some non-solutions.
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Last edited by v8archie; January 23rd, 2018 at 04:14 PM.
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January 23rd, 2018, 09:10 PM   #4
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You presumably didn't put in a constant of integration when doing the first integration.
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January 24th, 2018, 03:34 AM   #5
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Oh, you may be correct.
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March 6th, 2018, 01:00 PM   #6
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Since x does not appear explicitly in this equation, let u= dy/dx. Then $\displaystyle d^2y/dx^2= \frac{du}{dx}= \frac{du}{dy}\frac{dy}{dx}= u\frac{du}{dy}$.

The equation becomes $\displaystyle 2u\frac{du}{dy}= 3y^2$ which we can separate as $\displaystyle 2udu= 3y^2dy$. Integrating, $\displaystyle u^2= y^3+ C_1$.

$\displaystyle \frac{dy}{dx}= \pm\sqrt{y^3+ C_1}$.

That separates as $\displaystyle \frac{dy}{\sqrt{y^3+ C_1}}= dx$
and the problem is now to integrate that!
Thanks from topsquark

Last edited by skipjack; March 6th, 2018 at 09:21 PM.
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