# Power Series Solutions of Differential Equations

## Key Questions

• Assuming you know how to find a power series solution for a linear differential equation around the point ${x}_{0}$, you just have to expand the source term into a Taylor series around ${x}_{0}$ and proceed as usual.

This may add considerable effort to the solution and if the power series solution can be identified as an elementary function, it's generally easier to just solve the homogeneous equation and use either the method of undetermined coefficients or the method of variation of parameters.

• The solution is

y=c_0sum_{n=0}^infty{x^n}/{n!}=c_0e^x,

where ${c}_{0}$ is any constant.

Let us look at some details.

Let
$y = {\sum}_{n = 0}^{\infty} {c}_{n} {x}^{n}$
$y ' = {\sum}_{n = 1}^{\infty} n {c}_{n} {x}^{n - 1} = {\sum}_{n = 0}^{\infty} \left(n + 1\right) {c}_{n + 1} {x}^{n}$

So, we can rewrite $y ' - y = 0$ as

${\sum}_{n = 0}^{\infty} \left(n + 1\right) {c}_{n + 1} {x}^{n} - {\sum}_{n = 0}^{\infty} {c}_{n} {x}^{n} = 0$

by combining the summations,

$R i g h t a r r o w {\sum}_{n = 0}^{\infty} \left[\left(n + 1\right) {c}_{n + 1} - {c}_{n}\right] {x}^{n} = 0$

so, we have

$\left(n + 1\right) {c}_{n + 1} - {c}_{n} = 0 R i g h t a r r o w {c}_{n + 1} = \frac{1}{n + 1} {c}_{n}$

Let us observe the first few terms.

c_1=1/1c_0=1/{1!}c_0

c_2=1/2c_1=1/{2}cdot1/{1!}c_0=1/{2!}c_0

c_3=1/3c_2=1/3cdot1/{2!}c_0=1/{3!}c_0
.
.
.
c_n=1/{n!}c_0

Hence, the solution is

y=sum_{n=0}^infty1/{n!}c_0x^n=c_0sum_{n=0}^infty{x^n}/{n!}=c_0e^x,

where ${c}_{0}$ is any constant.

See below

#### Explanation:

Assuming a power series solution like this:

• $y = {a}_{0} + {a}_{1} x + {a}_{2} {x}^{2} + {a}_{3} {x}^{3} + \ldots = {\sum}_{0}^{\infty} {a}_{n} {x}^{n}$

• $\implies {y}^{'} = {\sum}_{1}^{\infty} n {a}_{n} {x}^{n - 1} q \quad q \quad {y}^{' '} = {\sum}_{2}^{\infty} n \left(n - 1\right) {a}_{n} {x}^{n - 2}$

With this power series:

$y ' ' + 2 x y ' + y = 0$

$\implies {\underbrace{{\sum}_{2}^{\infty} n \left(n - 1\right) {a}_{n} {x}^{n - 2}}}_{= {\sum}_{0}^{\infty} \left(m + 2\right) \left(m + 1\right) {a}_{m + 2} {x}^{m}} + 2 x {\sum}_{1}^{\infty} n {a}_{n} {x}^{n - 1} + {\sum}_{0}^{\infty} {a}_{n} {x}^{n} = 0$

$\implies {\sum}_{0}^{\infty} \left(n + 2\right) \left(n + 1\right) {a}_{n + 2} {x}^{n} + 2 {\sum}_{1}^{\infty} n {a}_{n} {x}^{n} + {\sum}_{0}^{\infty} {a}_{n} {x}^{n} = 0$

$\implies 2 {a}_{2} + {\sum}_{1}^{\infty} \left(n + 2\right) \left(n + 1\right) {a}_{n + 2} {x}^{n} + 2 {\sum}_{1}^{\infty} n {a}_{n} {x}^{n} + {a}_{0} + {\sum}_{1}^{\infty} {a}_{n} {x}^{n} = 0$

$\therefore {\underbrace{2 {a}_{2} + {a}_{0}}}_{= 0} + {\sum}_{1}^{\infty} \left({\underbrace{\left(n + 2\right) \left(n + 1\right) {a}_{n + 2} + \left(2 n + 1\right) {a}_{n}}}_{= 0}\right) {x}^{n} = 0$

This insists that the coefficient of the ${x}^{0}$ term is zero, but also that the coefficient of every ${x}^{i > 0}$ term is also zero, as the homogeneous DE requires.

The recurrence relation is:

• ${a}_{n + 2} = - \frac{\left(2 n + 1\right)}{\left(n + 2\right) \left(n + 1\right)} {a}_{n}$

This suggest 2 independent solutions to the DE, one for odd terms and one for even. These can be linearly super-imposed to reach a general null solution.

Even terms:

Setting: ${a}_{0} = 1 , {a}_{1} = 0$:

{( a_0 = 1),(a_2 = - 1/2 ),(a_4 = - 5/(4*3)* - 1/2 = 5/(4!) ),(a_6 = -(9)/(6*5) * 5/(4!) = - (5*9)/(6!) ),(a_8 = -(13)/(8*7) * - (5*9)/(6!) = (5*9*13)/(8!) ):}

 implies underbrace(1/(1!) x^0)_(k = 0) + underbrace((-1)^(1) 1/(2!) x^2)\_(k = 1) + (-1)^(2) 5/(4!) x^4 + (-1)^(3) (5*9)/(6!) x^8 + (-1)^(4) (5*9*13)/(8!) x^8 + cdots + underbrace((-1)^(k) (5* 9 * ... * (4 k - 3))/((2k)!) x^(2 k))_("but not for first 2 terms") + cdots = 0

An even solution is therefore:

y_E = 1 - x^2/2 + sum_2^oo (-1)^(k) (5* 9 * ... * (4 k - 3))/((2k)!) x^(2 k)

Odd terms:

Setting: ${a}_{0} = 0 , {a}_{1} = 1$, and copying the broad pattern:

{( a_1 = 1),(a_3 =- 3/(2*3) = - 3/(3!) ),(a_5 = - 7/(5*4)* - 3/(3!) = (3*7)/(5!) ),(a_7 = - 11/(7*6) * (3*7)/(5!) =- (3*7*11)/(7!)),(a_9 = -(3*7*11)/(7!)* -(15)/((9*8) )= (3*7*11*15)/(9!) ):}

 implies underbrace( x )_(k = 0) + underbrace((-1)^(1) 3/(3!) x^3)\_(k = 1) + (-1)^(2) (3*7)/(5!) x^5 + (-1)^(3) (3*7*11)/(7!) x^7 + cdots + underbrace((-1)^(k) (3* 7 * ... * (4 k -1))/((2k+1)!) x^((2 k+1)))_("but not for first term") + cdots = 0

An odd solution is therefore:

y_O =x + sum_1^oo (-1)^(k) (3* 7 * ... * (4 k -1))/((2k+1)!) x^((2 k+1))

Recognising the linearity:

$y = {c}_{1} \setminus {y}_{O} + {c}_{2} \setminus {y}_{E}$

So you have to add all that up

Finally , a screen grab from Socratic that always puts me of answering qu's like this in proper fashion:

I'd recommend: this

• Let us solve the differential equation $y ' ' = y$ by Power Series Method.

Let $y = {\sum}_{n = 0}^{\infty} {c}_{n} {x}^{n}$, where ${c}_{n}$ is to be determined.
By taking derivatives term by term,
$y ' = {\sum}_{n = 1}^{\infty} n {c}_{n} {x}^{n - 1}$
and
$y ' ' = {\sum}_{n = 2}^{\infty} n \left(n - 1\right) {c}_{n} {x}^{n - 2}$

So, $y ' ' = y$ becomes
${\sum}_{n = 2}^{\infty} n \left(n - 1\right) {c}_{n} {x}^{n - 2} = {\sum}_{n = 0}^{\infty} {c}_{n} {x}^{n}$
by shifting the indices on the summation on the left by 2,
${\sum}_{n = 0}^{\infty} \left(n + 2\right) \left(n + 1\right) {c}_{n + 2} {x}^{n} = {\sum}_{n = 0}^{\infty} {c}_{n} {x}^{n}$

By matching each coefficients,
(n+2)(n+1)c_{n+2}=c_n Rightarrow c_{n+2}=c_n/{(n+2)(n+1)}

Let us observe the first few even terms,
c_2=1/{2cdot1}c_0=1/{2!}c_0
c_4=1/{4cdot3}c_2=1/{4cdot3}cdot1/{2!}c_0=1/{4!}c_0
.
.
.
c_{2n}=1/{(2n)!}c_0

Let us observe the first few odd terms,
c_3=1/{3cdot2}c_1=1/{3!}c_1
c_5=1/{5cdot4}c_3=1/{5cdot4}cdot1/{3!}c_1=1/{5!}c_1
.
.
.
c_{2n+1}=1/{(2n+1)!}c_1

Now, we can find the solution $y$.
$y = {\sum}_{n = 0}^{\infty} {c}_{n} {x}^{n}$
by separating even terms and odd terms,
$= {\sum}_{n = 0}^{\infty} {c}_{2 n} {x}^{2 n} + {\sum}_{n = 0}^{\infty} {c}_{2 n + 1} {x}^{2 n + 1}$
by using the formulas for ${c}_{2 n}$ and ${c}_{2 n + 1}$ above,
=c_0sum_{n=0}^inftyx^{2n}/{(2n)!}+c_1sum_{n=0}^infty x^{2n+1}/{(2n+1)!}

Recall:
coshx=sum_{n=0}^infty x^{2n}/{(2n)!}
sinhx=sum_{n=0}^infty x^{2n+1}/{(2n+1)!}

Hence, $y = {c}_{0} \cosh x + {c}_{1} \sinh x$, where ${c}_{0}$ and ${c}_{1}$ are any constants.