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How do you use power series to solve the differential equation $y' = xy$?

seo-qna
Last updated date: 17th Jul 2024
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Answer
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Hint: In the above question, the concept is based on the concept of differentiating an equation using power series. The main approach towards solving this is to find a power series solution for the differential equation around point \[{x_0}\].

Complete step by step solution:
In mathematics, a power series is an infinite series of the form
\[\sum\limits_{n = 0}^\infty {{a_n}{{\left( {x - c} \right)}^n} = {a_0} + {a_1}{{\left( {x - c} \right)}^1} + {a_2}{{\left( {x - c} \right)}^2} + ...} \]
where ${a_n}$ is the coefficient of the nth term, c is a constant term and x is the variable.
Generally, it can be written as
Let
\[y = \sum\limits_{n = 1}^\infty {{c_n}{x^n}} \]
So now by derivating the above term we get,
\[y' = \sum\limits_{n = 1}^\infty {n{c_n}{x^{n - 1}}} \]
Now, the given differential equation is
$y' = xy$
By substituting the power series we get,
$\sum\limits_{n = 1}^\infty {n{c_{_n}}{x^{n - 1}} = x\sum\limits_{n = 0}^\infty {n{c_n}{x^n}} } $
By extracting the first term from the summation on the left,
\[{c_1} + \sum\limits_{n = 0}^\infty {n{c_n}{x^{n - 1}} = \sum\limits_{n = 0}^\infty {{c_n}{x^{n + 1}}} }
\]
By shifting the indices of the summation on the left by 2
\[{c_1} + \sum\limits_{n = 0}^\infty {\left( {n + 2} \right){c_{n + 2}}{x^{n + 1}} = \sum\limits_{n = 0}^\infty {{c_n}{x^{n + 1}}} } \]
By making the coefficients match,
\[{c_1} = 0\] and \[\left( {n + 2} \right){c_{n + 2}} = {c_n} \Rightarrow {c_{n + 2}} = \dfrac{{{c_n}}}{{n + 2}}\]
Now let us focus on the odd terms
\[
{c_3} = \dfrac{{{c_1}}}{3} = \dfrac{0}{3} = 0 \\
{c_5} = \dfrac{{{c_3}}}{5} = \dfrac{0}{5} = 0 \\
\]
The general term of the above odd series can be written as \[{c_{2n + 1}} = 0\]
Now let us look at the even terms.
$
{c_2} = \dfrac{{{c_0}}}{2} \\
{c_4} = \dfrac{{{c_2}}}{2} = \dfrac{{{c_0}}}{{4 \times 2}} = \dfrac{{{c_0}}}{{{2^2} \times 3!}} \\
$
The general term of the above even series can be written as
\[{c_{2n}} = \dfrac{{{c_0}}}{{{2^n} \times n!}}\] Therefore, by substituting we get,
\[y = \sum\limits_{n = 0}^\infty {\dfrac{{{c_0}}}{{{2^n} \times n!}}{x^{2n}} = {c_0}\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {\dfrac{{{x^2}}}{2}} \right)}^n}}}{{n!}} = {c_0}{e^{\dfrac{{{x^2}}}{2}}}} } \]
Now replacing x by \[\dfrac{{{x^2}}}{2}\]in \[{e^x}\] we get,
\[\sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}}} \]


Note: An important thing to note is that the general terms \[{c_{2n + 1}} = 0\] and \[{c_{2n}} = \dfrac{{{c_0}}}{{{2^n} \times n!}}\] is figured out in such a way that we need to check the pattern of the series of the odd terms and the even terms.Since in the odd terms were giving value as 0 and for even the power with base 2 is same that is why we get the above general terms.