Question

How do you solve power series to solve the Power series to solve the differential equation $y''+2xy'+y=0$?

Answer 1

Now we are given with a differential equation $y''+2xy'+y=0$ . To solve the equation by differential equation we will first consider the expansion of power series $y=\sum\limits_{1}^{\infty }{{{a}_{n}}{{x}^{n}}}$ . Now we will differentiate the equation to find y’ and y’’ and hence substitute the values in the given equation. Now we will simplify the equation by making the limits common. Now we will solve the equation to find a recurrence relation. Now with the recurrence relation we will get all the variables ${{a}_{n}}$ and hence the function y as $y=\sum\limits_{1}^{\infty }{{{a}_{n}}{{x}^{n}}}$

Complete step-by-step answer:
Now let us say $y=\sum\limits_{0}^{\infty }{{{a}_{n}}{{x}^{n}}}$ be power series expansion of the function y.
Now differentiating we get $y'=\sum\limits_{1}^{\infty }{n{{a}_{n}}{{x}^{n-1}}}$ and $y''=\sum\limits_{2}^{\infty }{n\left( n-1 \right){{a}_{n}}{{x}^{n-2}}}$ .
Now let us substitute the values in the given equation.
$\begin{aligned} & \Rightarrow y''+2xy'+2y=\sum\limits_{2}^{\infty }{n\left( n-1 \right){{a}_{n}}{{x}^{n-2}}}+2x\sum\limits_{1}^{\infty }{n{{a}_{n}}{{x}^{n-1}}}+\sum\limits_{0}^{\infty }{{{a}_{n}}{{x}^{n}}} \\\ & \Rightarrow y''+2xy'+2y=\sum\limits_{2}^{\infty }{n\left( n-1 \right){{a}_{n}}{{x}^{n-2}}}+2\sum\limits_{1}^{\infty }{n{{a}_{n}}{{x}^{n}}}+\sum\limits_{0}^{\infty }{{{a}_{n}}{{x}^{n}}} \\\ \end{aligned}$
Now substituting n = n + 2 to change the limits in the first summation we get,
$\Rightarrow y''+2xy'+2y=\sum\limits_{0}^{\infty }{\left( n+2 \right)\left( n+1 \right){{a}_{n+2}}{{x}^{n}}}+2\sum\limits_{1}^{\infty }{n{{a}_{n}}{{x}^{n}}}+\sum\limits_{0}^{\infty }{{{a}_{n}}{{x}^{n}}}$
Now let us separate the first term from the first and last summation. Hence we get,
$\Rightarrow y''+2xy'+2y=2{{a}_{2}}+\sum\limits_{1}^{\infty }{\left( n+2 \right)\left( n+1 \right){{a}_{n+2}}{{x}^{n}}}+2\sum\limits_{1}^{\infty }{n{{a}_{n}}{{x}^{n}}}+{{a}_{0}}+\sum\limits_{1}^{\infty }{{{a}_{n}}{{x}^{n}}}$

& \Rightarrow y''+2xy'+2y=2{{a}_{2}}+{{a}_{1}}+\sum\limits_{1}^{\infty }{\left( \left( n+2 \right)\left( n+1 \right){{a}_{n+2}}+2n{{a}_{n}}+{{a}_{n}} \right){{x}^{n}}} \\\ & \Rightarrow 0=2{{a}_{2}}+{{a}_{1}}+\sum\limits_{1}^{\infty }{\left( \left( n+2 \right)\left( n+1 \right){{a}_{n+2}}+2n{{a}_{n}}+{{a}_{n}} \right){{x}^{n}}} \\\ \end{aligned}$$ $\Rightarrow 2{{a}_{2}}+{{a}_{1}}=0$ and $\left( \left( n+2 \right)\left( n+1 \right){{a}_{n+2}}+2n{{a}_{n}}+{{a}_{n}} \right)=0$ Hence we can say that ${{a}_{1}}=-2{{a}_{2}}$ and $$\left[ \left( n+1 \right)\left( n+2 \right) \right]{{a}_{n+2}}=-\left( 2n+1 \right){{a}_{n}}$$ $${{a}_{n+2}}=-\dfrac{\left( 2n+1 \right){{a}_{n}}}{\left( n+1 \right)\left( n+2 \right)}$$ and ${{a}_{1}}=-2{{a}_{2}}$ . Now we have a recurrence relation and hence we can write the whole equation in terms of Hence with this recurrence relation we can write the whole function as $y=\sum\limits_{1}^{\infty }{{{a}_{m}}{{x}^{m}}}$ . Now assume ${{a}_{1}}=0,{{a}_{0}}=1$ to find odd terms and ${{a}_{0}}=0,{{a}_{1}}=1$ to find even terms. Hence we get the solution of the differential equation by Power series method. **Note:** Now note that when we write $$0=2{{a}_{2}}+{{a}_{1}}+\sum\limits_{1}^{\infty }{\left( \left( n+2 \right)\left( n+1 \right){{a}_{n+2}}+2n{{a}_{n}}+{{a}_{n}} \right){{x}^{n}}}$$ the zero on the left represents a zero polynomial which is $0+0x+0{{x}^{2}}+0{{x}^{3}}+...$ Hence comparing the equations we get the coefficient of ${{x}^{n}}$ must be zero for all n. Also note that while taking two solutions to find odd and even terms note that the two solutions must be independent from each other which means it is not multiple of each other