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Given y'' - x^2y' + xy = 0 about (0,0)

a. Find two solutions of the given differential equation.
b. Solve the IVP if y(0) = 2, and y'(0) = 5

I am having trouble with the re-indexing and solving for ck's

1 Answer

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By "ck" I'm assuming you're referring to the series solution's coefficients, i.e.
y=\displaystyle\sum_(k\ge0)c_kx^k.

Differentiating gives


y'=\displaystyle\sum_(k\ge1)kc_kx^(k-1)

y''=\displaystyle\sum_(k\ge2)k(k-1)c_kx^(k-2)

Substituting into the ODE yields


\displaystyle\sum_(k\ge2)k(k-1)c_kx^(k-2)-x^2\sum_(k\ge1)kc_kx^(k-1)+x\sum_(k\ge0)c_kx^k=0

\displaystyle\sum_(k\ge2)k(k-1)c_kx^(k-2)-\sum_(k\ge1)kc_kx^(k+1)+\sum_(k\ge0)c_kx^(k+1)=0

By extracting the first term (
k=0) of the third series, you can consolidate the second and third series:


\displaystyle\sum_(k\ge0)c_kx^(k+1)=\sum_(k\ge1)c_kx^(k+1)+c_0x

Next, since the first series starts with a constant term while the other two start with a quadratic term, remove the first two terms:


\displaystyle\sum_(k\ge2)k(k-1)c_kx^(k-2)=\sum_(k\ge4)k(k-1)c_kx^(k-2)+2c_2+6c_3x

Now shift the index so that the series begins at the same starting index by replacing
k with
k+3.


\displaystyle\sum_(k\ge4)k(k-1)c_kx^(k-2)=\sum_(k\ge1)(k+3)(k+2)c_(k+3)x^(k+1)

So now the series can be joined into one:


\displaystyle\sum_(k\ge1)\bigg[(k+3)(k+2)c_(k+3)-kc_k+c_k\bigg]x^(k+1)

which means the ODE is equivalent to


\displaystyle\sum_(k\ge1)\bigg[(k+3)(k+2)c_(k+3)-(k-1)c_k\bigg]x^(k+1)+c_0x+2c_2+6c_3x=0

It follows that
2c_2=0\implies c_2=0 and
c_0+6c_3=0. Using this, you solve the recurrence


(k+3)(k+2)c_(k+3)-(k-1)c_k=0\iff c_(k+3)=(k-1)/((k+3)(k+2))c_k

with the initial values
c_0 and
c_1.

Because
c_2=0, it follows that
c_5=0,c_8=0,\ldots; in general,
c_(k=3\ell-1)=0 for all
\ell\ge1.

Also, because the RHS of the recurrence vanishes when
k=1, it follows that
c_4=0,c_7=0,\ldots; or generally,
c_(k=3\ell+1)=0 for
\ell\ge1, leaving you with only one not-necessarily-zero term of
c_1.

So the hard part is finding a solution for
c_k when
k is a multiple of 3. By finding the first few of these terms, you'll start to notice a pattern. Starting with
c_0, you find


c_3=-\frac1{3*2}c_0

c_6=\frac2{6*5}c_3=-\frac1{6*5*3}c_0

c_9=\frac5{9*8}c_6=-\frac1{9*8*6*3}c_0

c_(12)=\frac9{12*11}c_9=-\frac1{12*11*9*6*3}c_0

and so on, following a general pattern of


c_(k=3\ell)=-(c_0)/((3\ell-1)\prod\limits_(i=1)^\ell 3i)

c_(k=3\ell)=-(c_0)/(3^\ell\ell!(3\ell-1))

valid for
\ell\ge1.

So the series solution is


y=\displaystyle\sum_(k\ge0)c_kx^k

y=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5+c_6x^6+\cdots

y=c_0+c_1x+\displaystyle\sum_(\ell\ge1)c_(3\ell-1)x^(3\ell-1)+\sum_(\ell\ge1)c_(3\ell)x^(3\ell)+\sum_(\ell\ge1)c_(3k+1)x^(3k+1)

y=c_0+c_1x+\displaystyle\sum_(\ell\ge1)c_(3\ell)x^(3\ell)

y=c_0+c_1x-c_0\displaystyle\sum_(\ell\ge1)(x^(3\ell))/(3^\ell\ell!(3\ell-1))

y=c_0+c_1x+c_0\displaystyle\sum_(\ell\ge1)(x^(3\ell))/(3^\ell\ell!(1-3\ell))

y=c_1x+c_0\displaystyle\sum_(\ell\ge0)(x^(3\ell))/(3^\ell\ell!(1-3\ell))

So the two solutions are
y_1=x and
y_2=\displaystyle\sum_(\ell\ge0)(x^(3\ell))/(3^\ell\ell!(1-3\ell)).

To solve the IVP, notice that when
x=0, you have


y(0)=\displaystyle\sum_(k\ge0)c_kx^k=c_0=2

y'(0)=\displaystyle\sum_(k\ge1)kc_kx^(k-1)=c_1=5

so the solution to the IVP is the same as that to the general solution with the unknowns replaced accordingly.
User Nick Urban
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