Question

How do you use a power series to solve the differential equation ${{y}^{''}}=y$?

Answer 1

In this problem we need to calculate the solution of the differential equation by using the power series. In this power series we will assume the solution as $y=\sum\limits_{n=0}^{\infty }{{{c}_{n}}{{x}^{n}}}$ where ${{c}_{n}}$ is to be determined. Now we will calculate all the values that we have in the given equation from the value of $y$. That means we will calculate the values like ${{y}^{'}}$, ${{y}^{''}}$ and so on by derivating the value of $y$ with respect to $x$. After calculating all the values, we will substitute them in the given equation and simplify the equation to the value of ${{c}_{n}}$. From the value of ${{c}_{n}}$ we can write the solution of the differential equation.

Complete step by step answer:
The differential equation is ${{y}^{''}}=y$.
Let us assume the solution of the given differential equation from the power series
$y=\sum\limits_{n=0}^{\infty }{{{c}_{n}}{{x}^{n}}}$
In the given equation we have the values $y$, ${{y}^{''}}$. To calculate the value of ${{y}^{''}}$ we need to derive the value of $y$ with respect to $x$ two times.
Differentiating the $y$ value with respect to $x$, then we will get
$\Rightarrow {{y}^{'}}=\sum\limits_{n=1}^{\infty }{n{{c}_{n}}{{x}^{n-1}}}$
Again, differentiating the above value with respect to $x$, then we will have
$\Rightarrow {{y}^{''}}=\sum\limits_{n=2}^{\infty }{n\left( n-1 \right){{c}_{n}}{{x}^{n-2}}}$
We have the given equation ${{y}^{''}}=y$. Substituting the values, we have in the given equation, then we will get
$\begin{aligned} & \Rightarrow {{y}^{''}}=y \\\ & \Rightarrow \sum\limits_{n=2}^{\infty }{n\left( n-1 \right){{c}_{n}}{{x}^{n-2}}}=\sum\limits_{n=0}^{\infty }{{{c}_{n}}{{x}^{n}}} \\\ \end{aligned}$
Shifting the indices on the summation on the left side by $2$, then we will get
$\Rightarrow \sum\limits_{n=0}^{\infty }{\left( n+2 \right)\left( n+1 \right){{c}_{n+2}}{{x}^{n}}}=\sum\limits_{n=0}^{\infty }{{{c}_{n}}{{x}^{n}}}$
Equating the both the values in the above equation, then we have
$\begin{aligned} & \Rightarrow \left( n+2 \right)\left( n+1 \right){{c}_{n+2}}={{c}_{n}} \\\ & \Rightarrow {{c}_{n+2}}=\dfrac{{{c}_{n}}}{\left( n+2 \right)\left( n+1 \right)} \\\ \end{aligned}$
Let us observe the first few even numbers according to the above condition by substituting $n=0,2,...,2n$, then we will get
$\Rightarrow {{c}_{2}}=\dfrac{1}{2.1}{{c}_{0}}=\dfrac{1}{2!}{{c}_{0}}$
$\Rightarrow {{c}_{4}}=\dfrac{1}{4.3}{{c}_{2}}=\dfrac{1}{4.3}.\dfrac{1}{2!}{{c}_{0}}=\dfrac{1}{4!}{{c}_{0}}$
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$\Rightarrow {{c}_{2n}}=\dfrac{1}{\left( 2n \right)!}{{c}_{0}}$
Now observe the few first odd terms based on above condition by substituting $n=1,3,...,2n+1$ then we will get
$\Rightarrow {{c}_{3}}=\dfrac{1}{3.2}{{c}_{1}}=\dfrac{1}{3!}{{c}_{1}}$
$\Rightarrow {{c}_{5}}=\dfrac{1}{5.4}{{c}_{3}}=\dfrac{1}{5.4}.\dfrac{1}{3!}{{c}_{1}}=\dfrac{1}{5!}{{c}_{1}}$
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$\Rightarrow {{c}_{2n+1}}=\dfrac{1}{\left( 2n+1 \right)!}{{c}_{1}}$
Now substituting these values in the solution which is $y=\sum\limits_{n=0}^{\infty }{{{c}_{n}}{{x}^{n}}}$, then we will get
$\Rightarrow y=\sum\limits_{n=0}^{\infty }{{{c}_{n}}{{x}^{n}}}$
Separating even terms and odd terms in the summation, then we will get
$\Rightarrow y=\sum\limits_{n=0}^{\infty }{{{c}_{2n}}{{x}^{2n}}}+\sum\limits_{n=0}^{\infty }{{{c}_{2n+1}}{{x}^{2n+1}}}$
Substituting the calculated values in the above equation, then we will get
$\Rightarrow y={{c}_{0}}\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{2n}}}{\left( 2n \right)!}}+{{c}_{1}}\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{2n+1}}}{\left( 2n+1 \right)!}}$
Hence the solution of the given differential equation ${{y}^{''}}=y$ is $y={{c}_{0}}\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{2n}}}{\left( 2n \right)!}}+{{c}_{1}}\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{2n+1}}}{\left( 2n+1 \right)!}}$ where ${{c}_{0}}$, ${{c}_{1}}$ are the constants.

Note: We can also use the hyperbolic formulas that we have in the solution. Substituting the formulas $\cosh x=\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{2n}}}{\left( 2n \right)!}}$, $\sinh x=\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{2n+1}}}{\left( 2n+1 \right)!}}$ in the solution then we will get
$y={{c}_{0}}\cosh x+{{c}_{1}}\sinh x$.
The above solution is easy to remember and simple to use in further problems.