# Thread: Need serious help understanding power series!

1. ## Need serious help understanding power series!

I don't understand this at all, and the professor is no help. I have no idea where to start. Please help!

(a) Show that y=ln(1+x) is the solution to the IVP (1+x)y"+y'=0, y(0)=0, y'(0)=1 for x>-1. Then find a power series solution around x0=0, and conclude that this power series is equal to ln(1+x). What is the radius of convergence?

(b) Find a power series solution (around x0=0) to the IVP (1+x)y"+2y'=0, y(0)=1, y'(0)=-1
What is the sum of this power series?

2. ## Re: Need serious help understanding power series!

1) Differentiate $\ln(1+x)$ twice and substitute in to show that the equation is satisfied. Also verify that it satisfies the initial conditions. To get the power series solution, you write $y=\sum_{n=0}^{\infty} a_nx^n$. You can differentiate this twice and use the initial conditions to get values for $a_0$ and $a_1$. You can also substitute the series and its derivatives into the equation to find values for other $a_n$, at first as a recurrance relation and then in closed form. Since both the series and $\ln(1+x)$ are solutions we can say that they are equal because of the uniqueness of solutions to linear equations.

3. ## Re: Need serious help understanding power series!

2) Is similar in getting the power series. To sat what the series sums to, you must solve the equation. To do that you can write $v(x)=y'(x)$ and solve the resulting first order linear and separable equation.

4. ## Re: Need serious help understanding power series!

I figured out the first part, thanks! If I differentiated it correctly, I got y'=∑an nxn-1 and y"=∑an n(n-1)xn-2 (I can't get the symbols to input into this correctly) After that, where do I input the initial conditions? And I don't know what recurrence relation or closed form means. sorry. Will this also give me the radius of convergence?

5. ## Re: Need serious help understanding power series!

For part b, I think I messed up somewhere. this is what I did:

(1+x)d2y/dx2+2dy/dx=0 v=dy/dx

(1+x)dv/dx+2v=0

dv/dx=-2v/(1+x) (÷v, then integrate)

ln(v)=-2ln(1+x)

v=1/(1+x)2=dy/dx

y=-1/(1+x)

The initial conditions, for y(0)=1 I get an answer of -1, and for y'(0)=-1, I get an answer of 1. They're backwards somehow.

6. ## Re: Need serious help understanding power series!

$(1+x)y'' +2y' = 0$

$\dfrac{y''}{y'} = -\dfrac{2}{1+x}$

$\ln|y'| = -2\ln|1+x| + C$

$y'(0) = -1 \implies \ln|-1| = -2\ln(1) + C \implies C = 0$

$y' = \dfrac{1}{(1+x)^2}$

$y = C - \dfrac{1}{1+x}$

$y(0) = 1 \implies C = 2$

$y = 2 - \dfrac{1}{1+x}$

check ...

$y' = \dfrac{1}{(1+x)^2}$

$y'' = -\dfrac{2}{(1+x)^3}$

original DE ... $(1+x)y'' +2y' = 0$

$(1+x) \cdot \bigg[-\dfrac{2}{(1+x)^3}\bigg] + 2 \cdot \dfrac{1}{(1+x)^2} = -\dfrac{2}{(1+x)^2} + \dfrac{2}{(1+x)^2} = 0$

7. ## Re: Need serious help understanding power series!

Use the initial conditions like this:
\begin{align*} y(x) &= \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \ldots \\ y(0) = 1 &= a_0 \end{align*}
and similarly for $y'(0)= -1$ to fix $a_1$.

Then put the series for $y(x)$, $y'(x)$ and $y''(x)$ into the differential equation and compare coefficients of $x^n$. You can start with coeffients of $x^2$ if you like for which you will get an expression relating $a_2$ to $a_0$ and $a_1$ which you already know. You can then continue with coefficients of $x^3$, $x^4$, etc.. But if you need a general expression for the $a_n$ in terms of $n$, you will have to compare coefficients of $x^n$. This gives a relation between $a_n$, $a_{n-1}$ amd $a_{n-2}$. This might be something like the recurrence relation for the Fibonacci numbers $a_n=a_{n-1}+a_{n-2}$ which can be solved (using the values of $a_0$ and $a_1$) to get a closed form expression $a_n=f(n)$.