y'(x) + x = y(x)

im told y(x)= sum(C_n)(x^n) power series method

i understand how to work the method im just confused about what happens to x?

does it go to 0?does it already have regular a singular point?help,thanks.

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- May 10th 2009, 09:33 AMsheep99quick question, power series method.
y'(x) + x = y(x)

im told y(x)= sum(C_n)(x^n) power series method

i understand how to work the method im just confused about what happens to x?

does it go to 0?does it already have regular a singular point?help,thanks. - May 10th 2009, 10:37 AMCalculus26
See the attachment for the details but the x doesn't magically disappear

but the x term is incorporated into determining the coefficient of the x term once the power series and its derivative are plugged back into the differential equation

You'll notice c1-c0 = 0 considering the contstant term

then 2c2-c1+1 = 0 in considering the linear term this is where x comes in

There are no singularities in this DE - May 10th 2009, 12:28 PMHallsofIvy
This is just y"- y= -x. x= 0 is not a singular point at all!

Let $\displaystyle y= \sum_{n=0}^\infty C_n x^n$. Then $\displaystyle y"= \sum_{n=2}^\infty n(n-1)x^{n-2}$ so the equation becomes $\displaystyle \sum_{n=2}^\infty n(n-1)x^{n-2}- \sum_{n= 0}^\infty C_n x^n= -x$.

In order to be able to compare "like powers", change the indices. In the first sum, let j= n-2 so that n= j+2. We have $\displaystyle \sum_{j= 0}^\infty (j+2)(j+1)C_{j+2}x^j- \sum_{j=0}^\infty C_j x^j= \sum_{j=0}^\infty ((j+2)(j+1)C_{j+2}- C_j)x^j= -x= 0- 1(x)+ 0(x^2)+ ...$

Setting coefficients of like powers equal we have

j= 0: $\displaystyle 2C_2- C_0= 0$

j= 1: $\displaystyle 3(2)C_3- C_1= -1$

(**That's**where the "-x" on the right hand side comes in!)

j> 1: $\displaystyle (j+2)(j+1)C_{j+2}- C_j= 0$