Need Help Solving a 2nd Order Nonlinear Differential Equation!

Sep 2014
9
0
NM
Solve the 2nd order nonlinear differential equation, with initial conditions y(0)=0 and y'(0)=1

y''=2ay^3-(a+1)y with a within [0,1]

I am pretty much lost on how to go about solving this. It would be greatly appreciated if someone could point me in the right direction on this. Thanks!
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
I would probably multiply both sides by y' giving

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}\,\frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= \left[ 2a\,y^3 - \left( a + 1 \right) y \right] \frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}$

Now notice that on the left hand side, if we let $\displaystyle \begin{align*} u = \frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}$, we get $\displaystyle \begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}$, giving

$\displaystyle \begin{align*} u\,\frac{\mathrm{d}u}{\mathrm{d}x} &= \left[ 2a\,y^3 - \left( a + 1 \right) y \right] \frac{\mathrm{d}y}{\mathrm{d}x} \\ \int{ u\,\frac{\mathrm{d}u}{\mathrm{d}x} \, \mathrm{d}x} &= \int{ \left[ 2a\,y^3 - \left( a + 1 \right) y \right] \frac{\mathrm{d}y}{\mathrm{d}x} \, \mathrm{d}x} \\ \int{ u\,\mathrm{d}u} &= \int{ 2a\,y^3 - \left( a + 1 \right) y \, \mathrm{d}y } \\ \frac{u^2}{2} + C_1 &= \frac{a\,y^4}{4} - \frac{ \left( a + 1 \right) y^2}{2} + C_2 \\ u^2 &= \frac{a\,y^4}{2} - \left( a + 1 \right) y^2 + C \textrm{ where } C = 2 \left( C_2 - C_1 \right) \\ \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 &= \frac{a\,y^4}{2} - \left( a + 1 \right) y^2 + C \end{align*}$

Now we know that when $\displaystyle \begin{align*} x = 0 \end{align*}$ we have $\displaystyle \begin{align*} y = 0 \end{align*}$ and $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = 1 \end{align*}$, so substituting in gives

$\displaystyle \begin{align*} 1^2 &= \frac{a \cdot 0^4}{2} - \left( a + 1 \right) \cdot 0^2 + C \\ 1 &= C \end{align*}$

and thus

$\displaystyle \begin{align*} \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 &= \frac{a\,y^4}{2} - \left( a + 1 \right) y^2 + 1 \end{align*}$

and hopefully you can now solve this DE :)
 
Sep 2014
9
0
NM
Thank you so much! I will go ahead and give it a shot!
 

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
@ Prove It. You dropped a 2 when integrating \(\displaystyle y^3\).
 
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