# Series solution for ode by undetermined coefficients

1. Nov 5, 2009

### John 123

1. The problem statement, all variables and given/known data
Obtain the Taylor series solution up to and including order 3 of the following non linear ode
$$y'=x^2+\sin y,y(0)=\frac{\pi}{2}$$

2. Relevant equations
After substituting the power series form of sin(y) I get:
$$y'=x^2+(y-\frac{y^3}{3!}+\frac{y^5}{5!}-\frac{y^7}{7!}.....)$$

3. The attempt at a solution
We require a series solution of the form:
$$y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+....$$
Then
$$y'=a_1+2a_2x+3a_3x^2+4a_4x^3+...$$
Substituting y and y' in the ode and equating coefficients of like powers of x gives:
$$a_1=a_0-\frac{a_0^3}{6}$$
$$2a_2=a_1-\frac{a_0^2a_1}{2}$$
$$3a_3=1+a_2-\frac{a_0a_1^2}{2}-\frac{a_2a_0^2}{2}$$
Then expressing all the a's in terms of a_0 and using :
$$a_0=y(0)=\frac{\pi}{2}$$
I get a result completely different to the book answer and cannot see my error.
$$y=\frac{\pi}{2}+x+\frac{1}{6}x^2+$$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 5, 2009

### clamtrox

Could it be that a_1 = sin a_0 instead and you just forgot infinite minus two terms from the expression? I don't get the book answer though, quickly glancing it would seem to me that a_2 = 0.

3. Nov 5, 2009

### clamtrox

Perhaps it should be $$\pi / 2 + x + x^3/6$$ ?

4. Nov 5, 2009

### clamtrox

One more thing: it might be pretty hard to solve the way you are trying to. Instead you can be a bit tricky. You are given the initial condition for y(0). This allows you to determine a_0=y_0 instantly. Now it's easy to solve y'(0) from the DE, giving you again instantly a_1. Then you can simply differentiate the DE again, solving y''(0) and y'''(0) as well.

5. Nov 5, 2009

Hi Clamtrox