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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

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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

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



    Answer  :  

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



    Answer  :  

    The series is given by,

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

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



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