Existence of closed form solution of an ODEy″+4yy′+y3=0.

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kachnaemra · Accepted Answer

Step 1Taking&amp;nbsp;y′=y2p(y)&amp;nbsp;and&amp;nbsp;y=eξ&amp;nbsp;we arrive atppξ′+2p2+4p+1=0,which is separable. After integration and rearranging you’ll get that1=cy(2p+2+2)1+24(−2p+2−2)1−2Now I’ll take&amp;nbsp;y=1u, giving the equationu=c(−2ux′+2+2)1+24(2ux′+2−2)1−2.For equations of the formu=F(u′)the solution is given via the method of integration by differentiation in “Handbook of Exact Solutions for Ordinary Differential Equations” 2ed. by Polyanin and Zaitsev implicitly asu=F(s),&amp;nbsp;x=∫F(s)&#039;sds+c,which is not difficult to show.Your equation has the implicit solutiony=1c(−2s+2+2)−1−24(2s+2−2)−1+2,x=∫csdds[(−2s+2+2)1+24(2s+2−2)1−2]ds+c2.If I’m inputting the integral correctly in Wolframalpha it has a fairly long solution involving Hypergeometric functions, which I will leave for you to get. An implicit solution is more than one can usually ask for!

Jesse Gates · Accepted Answer

Step 1Just to give you an idea of the problem you face withy″+4yy′+y3=0switch variables and write the equation as−x″[x′]3+4yx′+y3=0Using p=x′−p′p3+4yp+y3=0This on can be integrated but it leads to the implicit equation22tanh−1(y2p+22)−log⁡(1+4y2p+2y4p2)=4log⁡{y}+CNo way to recover p=??? but this form could be useful for a numerical integration.

Existence of closed form solution of an ODE

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