# Find a power series solution for the initial value problem: y" − 2xy' + 3y = 0 ; y(0) = −1, y'(0) = 2 y" − 2xy' + 3y = 0 ; y(0) = −1, y'(0) = 2you must write out the first five nonzero...

Find a power series solution for the initial value problem: y" − 2xy' + 3y = 0 ; y(0) = −1, y'(0) = 2

y" − 2xy' + 3y = 0 ; y(0) = −1, y'(0) = 2

you must write out the first five nonzero terms of

the solution.

### 1 Answer | Add Yours

Let

`y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...`

So:

`y'=a_1+2a_2x+3a_3x^2+4a_4x^3+...`

`y''=2a_2+6a_3x+12a_4x^2+...`

`y(0)=-1` tells us that `a_0=-1` and

`y'(0)=2` tells us that `a_1=2`

Plugging our series into the differential equation gives us:

`(2a_2+6a_3x+12a_4x^2+...)-2x(2+2a_2x+3a_3x^2+...)+3(-1+2x+a_2x^2+a_3x^3+...)=0`

We look at the coefficients of `1,x,x^2,x^3,...` on the right and left hand sides and set them equal to each other:

`2a_2-3=0` so `a_2=3/2`

`6a_3x-4x+6x=0` , so `a_3=-1/3`

`12a_4x^2-4a_2x^2+3a_2x^2=0` , `12a_4-4(3/2)+3(3/2)=0`

so `a_4 = 1/8`

`20a_5x^3-6a_3x^3+3a_3x^3=0`

So: `20a_5 - 3(-1/3)=0`

`a_5=-1/20`

`y=-1+2x+3/2 x^2 -1/3 x^3 +1/8 x^4 - 1/20 x^5+...`