
Differential Equations Ordinary and Partial Differential Equations Math Forum 
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January 23rd, 2018, 02:42 AM  #1 
Newbie Joined: May 2017 From: Moscow Posts: 6 Thanks: 0  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 12:02 PM. 
January 23rd, 2018, 02:54 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra 
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 nonsolutions. Last edited by v8archie; January 23rd, 2018 at 03:14 PM. 
January 23rd, 2018, 08:10 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
You presumably didn't put in a constant of integration when doing the first integration.

January 24th, 2018, 02:34 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra 
Oh, you may be correct.

March 6th, 2018, 12:00 PM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,197 Thanks: 872 
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! Last edited by skipjack; March 6th, 2018 at 08:21 PM. 

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