MHB 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!

 

[chisigma]

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!

That is an ODE of the type $\displaystyle y^{\ ''} = f(y)$, where $\displaystyle f(y)= 2\ a\ y^{3} - (1 + a)\ y$. With a little arrangement it becomes...

$\displaystyle y^{\ '}\ d y^{\ '} = f(y)\ d y\ (1)$

... the solution of which is...

$\displaystyle (y^{\ '})^{2} = 2\ \int \{2\ a\ y^{3} - (1+a)\ y\}\ d y = a\ y^{4} - (1 +a) y^{2} + c_{1} \implies y^{\ '} = \sqrt{a\ y^{4} - (1 + a)\ y^{2} + c_{1}}\ (2)$

... and, taking into account the initial conditions, is...

$\displaystyle y^{\ '} = \sqrt {a\ y^{4} - (1 + a)\ y^{2} + 1}\ (3)$

In that way You have obtained an explicit expression for y' and You are almost half way ... 'almost' because the awkward is now as we shall see in the next post...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Thatis an ODE of the type $\displaystyle y^{\ ''} = f(y)$, where$\displaystyle f(y)= 2\ a\ y^{3} - (1 + a)\ y$. With a littlearrangement it becomes... $\displaystyle y^{\'}\ d y^{\ '} = f(y)\ d y\ (1)$ ... the solution ofwhich is...$\displaystyle (y^{\'})^{2} = 2\ \int \{2\ a\ y^{3} - (1+a)\ y\}\ d y = a\ y^{4} - (1 +a)y^{2} + c_{1} \implies y^{\ '} = \sqrt{a\ y^{4} - (1 + a)\ y^{2} +c_{1}}\ (2)$... and, taking intoaccount the initial conditions, is... $\displaystyle y^{\'} = \sqrt {a\ y^{4} - (1 + a)\ y^{2} + 1}\ (3)$In that way You haveobtained an explicit expression for y' and You are almost half way... 'almost' because the awkward is now as we shall see in the nextpost...

We thus come to theODE ...$\displaystyle y^{\'} = \sqrt {a\ y^{4} - (1 + a)\ y^{2} + 1},\ y(0)= 0\ (1)$The standard procedure leads to the solution of the integral ...$\displaystyle \int\frac{d y} {\sqrt {a\ y^{4} - (1 + a)\ y^{2} + 1}}\ (2)$... which belongs to the family of elliptic integrals of the first type... integrate 1'/'sqrt '['a y'^'4 - '('1 '+' a')' y'^'2 '+' 1']' - Wolfram|Alpha++An alternative isthe search of a solution analitic in x=0, i.e. of the type... $\displaystyle y =\sum_ {n=0}^{\infty} a_{n}\ x^{n}\ (3)$The initial conditions give $a_{0}=0$ and $a_{1}=1$. The other $a_{n}$ can be obtained deriving the (1) ... $\displaystyle y^{\ ''} = y^{\ '} \frac{2 a y^{3} - (1 + a)\ y}{\{a\ y^4 - (1 + a)\ y^{2}+ 1\}^{\frac{1}{2}}} \implies a_{2}= 0\ (4)$

$\displaystyle y^{\ '''} = (y^{\ '})^{2} \ \{ \frac{6 a y^{2} - (1 + a)}{\{a\ y^4 - (1 + a)\ y^{2}+ 1\}^{\frac{1}{2}}} - \frac{(2 a y^{3} - (1 + a)\ y)^{2}}{\{a\ y^4 - (1 + a)\ y^{2}+ 1\}^{\frac{3}{2}}} \} + y^{\ ''}\ \text {other terms}\ \implies a_{3} = - \frac{1 + a}{6}\ (5) $

... so that an approximate solution is $\displaystyle y = x - \frac{1 + a}{6}\ x^{3} + ...$. Of course more terms can be computed, even if that is not a very comfortable task:( ...

Kind regards

$\chi$ $\sigma$