Question

Find the power series solution of the form y=a0+a1x+a2x^2+...+anx^n+... to the differential equations below

a) xy''+4y=0

b) y"'+y'+y=0

give the first six non zero terms in each solution.

thank you

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b) is y''+y'+y=0

Answer 1

For both ODEs, we take

`y=\displaystyle\sum_(n\ge0)a_nx^n`

which has derivatives

`y'=\displaystyle\sum_(n\ge0)na_nx^(n-1)=\sum_(n\ge0)(n+1)a_(n+1)x^n`

`y''=\displaystyle\sum_(n\ge0)n(n-1)a_nx^(n-2)=\sum_(n\ge0)(n+2)(n+1)a_(n+2)x^n`

Notice that we treat
`x^n` in such a way as to have
`a_0=y(0)`,
`a_1=y'(0)`, and
`a_2=y''(0)`

a.
`xy''+4y=0`

Substituting the relevant series above into the ODE gives

`\displaystyle\sum_(n\ge0)(n+2)(n+1)a_(n+2)x^(n+1)+4\sum_(n\ge0)a_nx^n=0`

`\displaystyle\sum_(n\ge1)n(n+1)a_(n+1)x^n+4\sum_(n\ge0)a_nx^n=0`

`\displaystyle\sum_(n\ge1)[n(n+1)a_(n+1)+4a_n]x^n+4a_0=0`

which immediately tells us
`4a_0=0`, or
`a_0=0`, and for every other coefficient
`(n\ge1)` we get

`a_(n+1)=-\frac4{n(n+1)}a_n`

So we have

`n=0\implies a_0=0`

`n=1\implies a_1=a_1`

`n=2\implies a_2=-(4a_1)/(2\cdot1)`

`n=3\implies a_3=-(4a_2)/(3\cdot2)=((-4)^2a_1)/(3\cdot2^2)`

`n=4\implies a_4=-(4a_3)/(4\cdot3)=((-4)^3a_1)/(4\cdot3^2\cdot2^2)`

and so on, with the general pattern

`a_n=((-4)^(n-1)a_1)/(n((n-1)!)^2)`

Then the first 6 non-zero terms of the solution are

`a_1\left(x-2x^2+\frac43x^3-\frac49x^4+\frac4{45}x^5-\frac8{675}x^6\right)`

# # #

b. I'm not sure if you mean to include the third derivative, or if the last part of your post is correcting it to be the second derivative. I'll assume the latter because it's slightly simpler.

`y''+y'+y=0`

Substituting the series above into the ODE gives

`\displaystyle\sum_(n\ge0)(n+2)(n+1)a_(n+2)x^n+\sum_(n\ge0)(n+1)a_(n+1)x^n+\sum_(n\ge0)a_nx^n=0`

`\displaystyle\sum_(n\ge0)[(n+2)(n+1)a_(n+2)+(n+1)a_(n+1)+a_n]x^n=0`

so that

`a_(n+2)=-(a_(n+1))/(n+2)-(a_n)/((n+2)(n+1))`

Finding a general pattern here is a bit more involved, so we'll compute the coefficients up to 6 non-zero terms:

`n=0\implies a_0=a_0`

`n=1\implies a_1=a_1`

`n=2\implies a_2=-\frac{a_0+a_1}2`

`n=3\implies a_3=\frac{a_0}6`

`n=4\implies a_4=(a_1)/(24)`

`n=5\implies a_5=-(a_0+a_1)/(120)`

which are all non-zero as long as
`y(0)\\eq0` and
`y'(0)\\eq0`, so the first 6 non-zero terms are

`a_0+a_1x-\frac{a_0+a_1}2x^2+\frac{a_0}6x^3+(a_1)/(24)x^4-(a_0+a_1)/(120)x^6`

# # #

Included below are plots of the exact solutions to both ODE (in blue) along with the first 6 terms found above, with assumed initial conditions.