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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 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, 12:09 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 Use W|A. January 23rd, 2018, 02:54 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 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 non-solutions. Thanks from Country Boy Last edited by v8archie; January 23rd, 2018 at 03:14 PM. January 23rd, 2018, 08:10 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 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: 7,674 Thanks: 2654 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: 3,264 Thanks: 902 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 08:21 PM. Tags 2y3y2 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

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