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Power Series Calculator
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Power series calculator helps to find out the solution of a differential equation in the form of $\sum_{n=0}^{\infty }a_{n}x^{n}$. There are some steps to do this calculation which is given in detail in the below section. The power series equation is given as,

f(x)=$\sum_{n=0}^{\infty }a_{n}x^{n}$

## Steps for Power Series Calculator

Step 1 :

Consider y=$\sum_{n=0}^{\infty }a_{n}x^{n}$ is a solution of the given function. Determine the values of y',y'' according to the equation.

Step 2 :

Substitute the above values in the equation, adjust the equation such a way that both sides have the same power of x.

Step 3 :

Evaluating the coefficient of like terms consisting power of x.

Step 4 :

Determine the values of a0, a1, a2 ........an

Step 5 :

Substitute the values in the given formula.

f(x)=$\sum_{n=0}^{\infty }a_{n}x^{n}$

## Problems on Power Series Calculator

1. ### Solve y'-xy=0 using power series.

Step 1 :

The given equation is, y'-xy=0

Let y=$\sum_{n=0}^{\infty }a_{n}x^{n}$ is a solution of the given function

y'=$\sum_{n=0}^{\infty }na_{n}x^{n-1}$

Step 2 :

$\sum_{n=0}^{\infty }na_{n}x^{n-1}$-x$\sum_{n=0}^{\infty }a_{n}x^{n}$=0

$\sum_{n=0}^{\infty }na_{n}x^{n-1}$=x$\sum_{n=0}^{\infty }a_{n}x^{n}$

$\sum_{n=0}^{\infty }(n+2)a_{n+2}x^{n+1}$=$\sum_{n=0}^{\infty }a_{n}x^{n+1}$

Step 3 :

(n+2)$a_{n+2}$=$a_{n}$

$a_{n+2}$=$\frac{a_{n}}{n+2}$

Step 4 :

$a_{2}$=$\frac{a_{0}}{2}$
$a_{3}$=$\frac{a_{1}}{3}$
$a_{4}$=$\frac{a_{2}}{4}$=$\frac{a_{0}}{2.4}$
$a_{2k}$=$\frac{a_{0}}{2^{k}k!}$
$a_{2k+1}$=$\frac{a_{1}}{3.5.7....(2k+1)}$

Step 5 :

y=$\sum_{n=0}^{\infty }a_{n}x^{n}$=$a_{0}\sum_{k=0}^{\infty }\frac{x^{2k}}{2^{k}k!}$+$a_{1}\sum_{k=0}^{\infty }\frac{x^{2k+1}}{3.5.7.....(2k+1)}$

The series is given by,

y=$a_{0}\sum_{k=0}^{\infty }\frac{x^{2k}}{2^{k}k!}$+$a_{1}\sum_{k=0}^{\infty }\frac{x^{2k+1}}{3.5.7.....(2k+1)}$

2. ### Solve the differential equation y'-5y=0 using power series expansion.

Step 1 :

The given equation is, y'-5y=0

Let y=$\sum_{n=0}^{\infty }a_{n}x^{n}$ is a solution of the given function

y'=$\sum_{n=0}^{\infty }na_{n}x^{n-1}$

Step 2 :

$\sum_{n=0}^{\infty }na_{n}x^{n-1}$-5$\sum_{n=0}^{\infty }a_{n}x^{n}$=0

$\sum_{n=0}^{\infty }na_{n}x^{n-1}$=5$\sum_{n=0}^{\infty }a_{n}x^{n}$

$\sum_{n=0}^{\infty }(n+1)a_{n+1}x^{n}$=5$\sum_{n=0}^{\infty }a_{n}x^{n}$

Step 3 :

(n+1)$a_{n+1}$=5$a_{n}$
$a_{n+1}$=$\frac{5a_{n}}{n+1}$

Step 4 :

$a_{1}$=5$a_{0}$

$a_{2}$=$\frac{5a_{1}}{2}$=$\frac{5^{2}a_{0}}{2!}$

$a_{3}$=$\frac{5a_{2}}{3}$=$\frac{5^{3}a_{0}}{3!}$

$a_{n}$=$\frac{5^{n}a_{0}}{n!}$

Step 5 :

y=$\sum_{n=0}^{\infty }a_{n}x^{n}$

y=$\sum_{n=0}^{\infty }\frac{5^{n}a_{0}}{n!}x^{n}$

y=$a_{0}\sum_{n=0}^{\infty }\frac{(5x)^{n}}{n!}$

The series is given by,

y=$a_{0}\sum_{n=0}^{\infty }\frac{(5x)^{n}}{n!}$

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