y''+y'=0 solve diff.equation using infinite series

Question

Solve the following differential equation y''+y'=0 using infinite series method and following these steps:

1)Assume there is a power series solution and rewrite the equation in power series form.

2)Adjust the summations appropiately for solving (adjust the "n").

3)Write the equation as a sigle summation.

4)Find a recursive relationship for the constant.

5)Generate the first 10 constants c0,c1,c2...,c9

6) Place constants in the original power series

7)IF POSSIBLE, find a closed form solution.

I attach a picture of how it should look the procedure. IMPORTANT NOTE: The one on the picture is not y"+y'=0, is y"-y'=0. I am asking for the solution of y"+y'=0.example picture link

Answer 1

You are told exactly what to do. Do it! "1)Assume there is a power series solution and rewrite the equation in power series form." Let $y= \sum_{n=0}^\infty a_nx^n$ Then $y'= \sum_{n=1}^\infty na_nx^{n-1}$ and $y''= \sum_{n= 2}^\infty n(n-1)a_nx^{n-2}$. (Note: the derivative of a constant is 0. That is why the first derivative starts at n= 1 and the second at n= 2. The n= 0 and and n= 1 terms are 0. The equation becomes $y''+ y'= \sum_{n=2}^\infty n(n-1)a_nx^{n-2}+ \sum_{n=1}^\infty na_nx^n= 0$.

"2)Adjust the summations appropriately for solving (adjust the "n")." We cannot combine the two sums because they have different powers of x for a given "n". To be able to match then change the indices in the two sums to give the same power of x. There are many different ways to do that. I would let j= n- 2 in the first sum and j= n- 1 in the second. In the first sum $a_n= a_{j+ 2}$ and when n= 2 j=0 so that sum becomes $\sum_{j=0}^\infty (j+2)(j+1)a_{j+1}x^j$. In the second sum $a_n= a_{j+1}$ and when n=1, j= 0 so that the sum becomes $\sum_{j=0}^\infty (j+ 1)a_{j+1}x^j$. The equation becomes $\sum_{j=0}^\infty (j+2)(j+1)a_{j+2}x^j+ \sum_{j=0}^\infty (j+ 1)a_{j+1}x^j= 0$

"3)Write the equation as a single summation." Since both sums have x to the j power and both start at j= 0 we can just add the coefficients: $\sum_{n=0}^\infty ((j+2)(j+1)a_{j+2}+ (j+1)a_j)x^j= 0$.

"4)Find a recursive relationship for the constant." In order that the previous equation be 0 for all x, all coefficients must be 0. That is $(j+2)(j+1)a_{j+2}+ (j+1)a_{j+1}= 0$ so $(j+2)(j+1)a_{j+2}= -(j+1)a_j$ and $a_{j+2}= -\frac{j+1}{(j+2)(j+1)}a_j= -\frac{1}{j+2}a_j$. That's the "recursive relation".

"5)Generate the first 10 constants c0,c1,c2...,c9" Those would be my $a_0$, $a_1$, etc. To determine values for $a_0$ and $a_1$ we would need "initial values". But given $a_0$ and $a_1$ we can set j= 0 to get $a_2= -\frac{1}{2}a_0$. Set j= 1 to get $a_3= -\frac{1}{3}a_1$. Set j= 2 to get $a_4= -\frac{1}{4}a_2= \frac{1}{8}a_0$, etc.

"6) Place constants in the original power series" So replace each $a_n$ with the values from (5). You will get some terms as multiples of $a_0$ and others as multiples of $a_1$. Separate those and write one as $a_0$ times a power series, the other as $a_1$ times a power series.

"7)IF POSSIBLE, find a closed form solution." You might be able to recognize those series as being power series for specific functions.

If not, you could always write z= y' so the differential equation become z'+ z= 0 so that z'= -z. Write that as dz/z= -dx. Integrating, ln|z|= -x+ C so $z= y'= C'e^{-x}$ and then $y= -C'e^{-x}+ D$.

Answer 2

I'll start at step $7$, if you don't mind.

$$y''+y'=0\implies y''=-y'\implies y'=C_1e^{-x}\implies y=C_2-C_1e^{-x}$$

So the solution is simply given by $y=C_2-C_1e^{-x}$.

From there, it is well known that

$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$

Answer 3

Suppose $y(x)=\sum_{n=0}^\infty c_nx^n$; then \begin{align} y'(x)&=\sum_{n=1}^{\infty}nc_nx^{n-1}\\[4px] y''(x)&=\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2} \end{align} With a change of indices, \begin{align} y'(x)&=\sum_{n=0}^{\infty}(n+1)c_{n+1}x^{n}\\[4px] y''(x)&=\sum_{n=0}^{\infty}(n+2)(n+1)c_{n+2}x^{n} \end{align} The equation now gives $$ (n+1)c_{n+1}=(n+2)(n+1)c_{n+2} $$ so $$ c_{n+2}=\frac{c_{n+1}}{n+2} $$ You have no control on $c_0$ and $c_1$, which can be arbitrary. Then $$ c_2=\frac{c_1}{2}, \quad c_3=\frac{c_2}{3}=\frac{c_1}{2\cdot 3}, \quad c_4=\frac{c_3}{4}=\frac{c_1}{2\cdot 3\cdot 4} $$ Can you spot a pattern?

Answer 4

$y'' -y' = 0$

$Suppose:y = \sum_\limits{n=0}^{\infty} c_nx^n\\ y''-y' = \sum_\limits{n=2}^{\infty} (n)(n-1)c_nx^{n-2} + \sum_\limits{n=1}^{\infty} (n)c_nx^{n-1}\\ \sum_\limits{n=0}^{\infty} (n+2)(n+1)c_{n+2}x^{n} + \sum_\limits{n=0}^{\infty} (n+1)c_{n+1}x^{n}\\ \sum_\limits{n=0}^{\infty} \left((n+2)(n+1)c_{n+2}-(n+1)c_{n+1}\right)x^{n} = 0\\ (n+2)(n+1)c_{n+2}-(n+1)c_{n+1}=0\\ c_{n+2}=\frac {c_{n+1}}{n+2}\\ c_2 = \frac {c_1}{2}\\ c_3 = \frac {c_1}{6}\\ c_n = \frac {c_1}{n!}\\ y = c_0 + c_1 \sum_\limits{n=0}^{\infty} \frac {x^n}{n!}\\\ y = c_0 + c_1 e^x$