Question

How do you use Power Series to solve the differential equation ${y}'-y=0~$?

Answer 1

In this problem, we need to calculate the solution of the differential equation by using the power series. For this, we will assume the solution as $y=\sum\limits_{n=0}^{\infty }{{{c}_{n}}{{x}^{n}}}$ where ${{c}_{n}}$ is to be determined. Next, we will calculate all the values that we have in the given equation from the value of $y$. That means we will calculate the value of ${y}'$ 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 solution:
From our problem, the differential equation is ${y}'-y=0~$.
Let us assume the solution of the given differential equation from the power series, that is,
$y=\sum\limits_{n=0}^{\infty }{{{c}_{n}}{{x}^{n}}}$
In the given equation, we have the values $y,{y}'$. To calculate the value of ${y}'$, we have to derivate the value of $y$ with respect to $x$.
Therefore, on differentiating the value of $y$ with respect to $x$, we get:
$\Rightarrow {y}'=\sum\limits_{n=1}^{\infty }{n{{c}_{n}}{{x}^{n-1}}}$
Now, we have our equation as ${y}'-y=0~$.
Therefore, substituting the values in our given equation, we get:
$\Rightarrow {y}'-y=0~$
$\Rightarrow \sum\limits_{n=1}^{\infty }{n{{c}_{n}}{{x}^{n-1}}}-\sum\limits_{n=0}^{\infty }{{{c}_{n}}{{x}^{n}}}=0$
Now, by extracting the first term from the summation on the left, we get:
$\Rightarrow \sum\limits_{n=0}^{\infty }{\left( n+1 \right){{c}_{n+1}}{{x}^{n}}}-\sum\limits_{n=0}^{\infty }{{{c}_{n}}{{x}^{n}}}=0$
Now, on combining the summation, we get:
$\Rightarrow \sum\limits_{n=0}^{\infty }{\left[ \left( n+1 \right){{c}_{n+1}}-{{c}_{n}} \right]}{{x}^{n}}=0$
Now, on making coefficient match, we get:
$\Rightarrow \left( n+1 \right){{c}_{n+1}}-{{c}_{n}}=0$
Now, on equating, we get:
$\Rightarrow {{c}_{n+1}}=\dfrac{1}{n+1}{{c}_{n}}$
Now, Let us observe the first few terms by substituting $n=0,1,2,...,n$.
Therefore, on substituting these value, we get:
$\Rightarrow {{c}_{1}}=\dfrac{1}{1}{{c}_{0}}=\dfrac{1}{1!}{{c}_{0}}$
$\Rightarrow {{c}_{2}}=\dfrac{1}{2}{{c}_{1}}=\dfrac{1}{2}\cdot \dfrac{1}{1!}{{c}_{0}}=\dfrac{1}{2!}{{c}_{0}}$
$\Rightarrow {{c}_{3}}=\dfrac{1}{3}{{c}_{2}}=\dfrac{1}{3}\cdot \dfrac{1}{1!}{{c}_{0}}=\dfrac{1}{3!}{{c}_{0}}$
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$\Rightarrow {{c}_{n}}=\dfrac{1}{n!}{{c}_{0}}$
Now, substituting these values in the solution which is $y=\sum\limits_{n=0}^{\infty }{{{c}_{n}}{{x}^{n}}}$.
Therefore, we get:
$\Rightarrow y=\sum\limits_{n=0}^{\infty }{{{c}_{n}}{{x}^{n}}}$
On substituting:
$\Rightarrow y=\sum\limits_{n=0}^{\infty }{\dfrac{1}{n!}{{c}_{0}}{{x}^{n}}}$
Therefore, on simplifying, we get:
$\Rightarrow y=\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{n}}}{n!}{{c}_{0}}}={{c}_{0}}{{e}^{x}}$
where ${{c}_{0}}$ is any constant.

Note: We can also use the hyperbolic formulas that we have in the solution. By substituting it in the formula, we can find $y={{c}_{0}}\cosh x+{{c}_{1}}\sinh x$, where ${{c}_{0}}$ and ${{c}_{1}}$ is any constant.
The above solution is easy to remember and simple to use.