# Need Help Solving a 2nd Order Nonlinear Differential Equation!

#### frank1234

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
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

#### frank1234

Thank you so much! I will go ahead and give it a shot!

#### Jester

MHF Helper
@ Prove It. You dropped a 2 when integrating $$\displaystyle y^3$$.

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