Non homogeneous ODE particular solution using power series

[cybla]

Homework Statement

Find the particular solution to the ODE y"+y=x using power series

Homework Equations

y=$\sum(a_{n}x^{n})$

The Attempt at a Solution

i tried plugging in y=$\sum(a_{n}x^{n})$ into the original equation and comparing coefficients of x to the first degree, but i am not sure how to really compare them.

 

[ehild]

Try to write out the power series: y=a₀+a₁x+a₂x²+a₃x³ ... Find the second derivative and plug back into the DE. Collect and compare the x terms on equal power, they must be equal on both sides of the equation.
Show your work.

ehild

 

[cybla]

Ok... so

y=a$_{0}$+a$_{1}$x+a$_{2}$x$^{2}$+a$_{3}$x$^{3}$+a$_{4}$x$^{4}$+a$_{5}$x$^{5}$...

y"= 2a$_{2}$+6a$_{3}$x+12a$_{4}$x$^{2}$+20a$_{5}$x$^{3}$+30a$_{6}$x$^{4}$+42a$_{7}$x$^{5}$...

x= 0+x+0x$^{2}$+0x$^{3}$+0x$^{4}$+0x$^{5}$...

y"+y=x

2a$_{2}$+6a$_{3}$x+12a$_{4}$x$^{2}$+20a$_{5}$x$^{3}$+30a$_{6}$x$^{4}$+42a$_{7}$x$^{5}$...+a$_{0}$+a$_{1}$x+a$_{2}$x$^{2}$+a$_{3}$x$^{3}$+a$_{4}$x$^{4}$+a$_{5}$x$^{5}$...
= 0+x+0x$^{2}$+0x$^{3}$+0x$^{4}$+0x$^{5}$...

a$_{0}$+2a$_{2}$=0
a$_{1}$+6a$_{3}$=1
a$_{2}$+12a$_{4}$=0

etc.

do i write a linear system of equations for 3 variables and solve it in order to find some of the a values? Or i should i use the fact from the undetermined coefficients method that the form must be y=A + Bx... and then solve?

 

[ehild]

You need to find one particular solution, so you have quite great freedom to choose the first a-s. And a₀ and a₁ determine all the other coefficients. Notice that choosing a₀=0 all other coefficients of even index are also zero. What happens if you choose a₁=1?

ehild

 

[cybla]

So the choice of a's (in this case a$_{0}$ and a$_{1}$) is up to me, but the final answer has to satisfy the DE, right?

 

[ehild]

The choice of the first two a-s, a₀ and a₁ is up to you, and all the others are obtained from them. The resultant power series is solution of the DE. But you need something simple, and an infinite power series is not that. So you may try to terminate it. If one of the a_2k coefficients is zero, then a_2(k-1)=0 and a_2(k+1)=0 so all the even index a-s are zero. The same with the a_2k+1 coefficients. If a_2k+1=0, a_2(k+1)+1=0 and so on, and also a_2k-1=0 so all the odd-index a-s are zero but the first one, a₁. That is you can find a partial solution which has only one term. What is it then?ehild

 

[cybla]

alright i understand. So for this problem if a$_{0}$=0 and a$_{1}$=1 then the solution is y=x

Thank you for your help

 

[ehild]

alright i understand. So for this problem if a$_{0}$=0 and a$_{1}$=1 then the solution is y=x

Thank you for your help

Yes, that simple! One could have seen at once, but using the power series was such a fun!

ehild