Question

How do you find a power series solution of a non-homogeneous differential equation?

Answer 1

To find the power series solution, we will begin by assuming a general form of the solution, that is, $y=\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{x}^{n}}}$. Then, as per the question’s requirement, we compute the derivative of the assumed solution. Then, we substitute the derivatives in the given differential and solve for the coefficients which have a recurrence relation. And having solved that, we use the Taylor series wherever required and get the required power series solution of the non-homogeneous differential equation.

Complete step-by-step answer:
According to the given question, we have to show how a power series solution of a non-homogeneous differential equation is found.
Non-homogeneous equations can be referred to as a differential equation having terms which does not include the dependent variable. Else it becomes a homogeneous differential equation.
For example - $xy'+2y=\sin x$, here $x$ is the independent variable and $y$ is the dependent variable. The term $\sin x$ does not have the dependent variable. Therefore, the equation is a non-homogeneous differential equation.
Power series solution refers to the series having certain coefficients which have recurrence relation and is used to write the general form of a differential equation’s solution.

When we are given a non-homogeneous differential equation, the power series solution is found first by assuming that the solution has a form similar to the expression,
$y=\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{x}^{n}}}$-----(1)
The above given equation (1) is the general form of the solution which can also be written as,
$y={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}+...$
Next step is computing the $i$th derivative of $y$ as per required in the question, we get the expression as,
${{y}^{i}}=\sum\limits_{n=0}^{\infty }{{{a}_{n}}(n-i+1)(n-i+2)...n{{x}^{n-1}}}$
${{y}^{i}}=\sum\limits_{n=i}^{\infty }{{{a}_{n}}(n-i+1)(n-i+2)...n{{x}^{n-1}}}$------(2)
We now have the derivatives of $y$as per required in the question, so we will substitute the equation (1) and (2) in the given differential equation. We will then get a recurrence relation for the coefficients ${{a}_{n}}$.
We then solve for the coefficients and we then use Taylor series around the constant and expand the coefficient function ${{a}_{n}}$ as applicable.
So, we have the power series solution of a non-homogeneous differential equation.
Let us understand this using an example,
The non-homogeneous differential equation we have is,
$y'+xy=1+x$-----(1)
Let the solution of the above expression be
$y=\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{x}^{n}}}$----(2)
Next, we will compute the first derivative of $y$, we get,
$y'=\sum\limits_{n=0}^{\infty }{n{{a}_{n}}{{x}^{n-1}}}$-----(3)
Now, we will substitute the equation (2) and (3), we get,
$\sum\limits_{n=0}^{\infty }{n{{a}_{n}}{{x}^{n-1}}}+x\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{x}^{n}}}=1+x$
$\Rightarrow \sum\limits_{n=0}^{\infty }{n{{a}_{n}}{{x}^{n-1}}}+\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{x}^{n+1}}}=1+x$
First thing we have to do is make the exponents of $x$ equal, so we will increase ‘n’ by 1 for the first term and decrease ‘n’ by 1 for the second term. We get,
$\sum\limits_{n=0}^{\infty }{(n+1){{a}_{n+1}}{{x}^{n}}}+\sum\limits_{n=1}^{\infty }{{{a}_{n-1}}{{x}^{n}}}=1+x$------(4)
Now, we will write the terms for each expression, so we have,
$\sum\limits_{n=0}^{\infty }{(n+1){{a}_{n+1}}{{x}^{n}}}=1{{a}_{1}}+2{{a}_{2}}x+\sum\limits_{n=2}^{\infty }{(n+1){{a}_{n+1}}{{x}^{n}}}$-----(5)
$\sum\limits_{n=1}^{\infty }{{{a}_{n-1}}{{x}^{n}}}={{a}_{0}}x+\sum\limits_{n=2}^{\infty }{{{a}_{n-1}}{{x}^{n}}}$-----(6)
Substituting the equation (5) and (6), we get,
$\Rightarrow 1{{a}_{1}}+2{{a}_{2}}x+\sum\limits_{n=2}^{\infty }{(n+1){{a}_{n+1}}{{x}^{n}}}+{{a}_{0}}x+\sum\limits_{n=2}^{\infty }{{{a}_{n-1}}{{x}^{n}}}=1+x$
On rearranging, we get,
$\Rightarrow 1{{a}_{1}}+2{{a}_{2}}x+{{a}_{0}}x+\sum\limits_{n=2}^{\infty }{((n+1){{a}_{n+1}}}+{{a}_{n-1}}){{x}^{n}}=1+x$ -----(7)
Comparing the LHS and RHS, we get,
${{a}_{1}}=1$
And $2{{a}_{2}}+{{a}_{0}}=1$
$\Rightarrow {{a}_{2}}=\dfrac{1-{{a}_{0}}}{2}$
Where ${{a}_{0}}$ is the arbitrary constant.
And the rest of the part of equation (7) equals to 0, we get,
$(n+1){{a}_{n+1}}+{{a}_{n-1}}=0$
$\Rightarrow {{a}_{n+1}}=\dfrac{-{{a}_{n-1}}}{n+1}$
Lets ${{a}_{0}}=1$
${{a}_{1}}=1$
${{a}_{2}}=\dfrac{1-{{a}_{0}}}{2}=0$
${{a}_{3}}=\dfrac{-{{a}_{1}}}{3}=\dfrac{-1}{3}$
${{a}_{4}}=\dfrac{-{{a}_{2}}}{4}=0$
${{a}_{5}}=\dfrac{-{{a}_{3}}}{5}=\dfrac{-1}{5.3}$
So, the sequence becomes,
$1,1,0,\dfrac{-1}{3},0,\dfrac{-1}{5.3},...$
Generalised form of the above sequence is,
${{a}_{n}}={{(-1)}^{n+1}}\dfrac{(2n-2)(2n-4)..4.2}{(2n-1)(2n-3)..5.3}={{(-1)}^{n+1}}\dfrac{(2n-2)!}{(2n-1)!}$
Substituting all the obtained values in the equation (2),
Therefore, we get the solution as,
$y={{a}_{0}}+\sum\limits_{n=1}^{\infty }{{{(-1)}^{n+1}}\dfrac{(2n-2)!}{(2n-1)!}{{x}^{2n}}}$

Note: The power series solution is quite big, so it should be done carefully and step wise. After substituting the derivatives in the given differential equation, the next step should always be equalizing the power of the independent variable. And for non-polynomial coefficient we will use Taylor series to expand their function and multiply them by Taylor series for their derivative.