Exercise:
Solve the
following D.E $y''=y'$ (by Power Series
)
Solution: we
will solve this D.E using power series expansion as follows :
Let $y=\sum\limits_{i=0}^{\infty
}{{{c}_{i}}{{x}^{i}}}$ then $y'=\sum\limits_{i=1}^{\infty
}{i{{c}_{i}}{{x}^{i-1}}}\,\,\And \,\,y''=\sum\limits_{i=2}^{\infty }{i\left(
i-1 \right){{c}_{i}}{{x}^{i-2}}}=\sum\limits_{i=0}^{\infty }{\left( i+2
\right)\left( i+1 \right){{c}_{i+2}}{{x}^{i}}}$
So $y''-y'=0\Rightarrow
\sum\limits_{i=0}^{\infty }{\left( i+2 \right)\left( i+1
\right){{c}_{i+2}}{{x}^{i}}-\sum\limits_{i=1}^{\infty
}{i\,{{c}_{i}}{{x}^{i-1}}}=0}$
Replace $i$ by
$i+1$ for the second term
So $\sum\limits_{i=1}^{\infty
}{i{{c}_{i}}{{x}^{i-1}}}=\sum\limits_{i=0}^{\infty }{\left( i+1
\right){{c}_{i+1}}{{x}^{i}}}$
Thus $y''-y'=0\Leftrightarrow
\sum\limits_{i=0}^{\infty }{\left( i+2 \right)\left( i+1
\right){{c}_{i+2}}{{x}^{i}}-\sum\limits_{i=0}^{\infty }{\left( i+1
\right){{c}_{i+1}}{{x}^{i}}=0}}$
$\Rightarrow
\sum\limits_{i=0}^{\infty }{\left[ \left( i+2 \right)\left( i+1
\right){{c}_{i+2}}-\left( i+1 \right){{c}_{i+1}} \right]{{x}^{i}}=0}$ $\Rightarrow
{{c}_{i+2}}=\frac{1}{i+2}{{c}_{i+1}}$
For $i=0\Rightarrow
{{c}_{2}}=\frac{1}{2}{{c}_{1}}$
For $i=1\Rightarrow
{{c}_{3}}=\frac{1}{3}{{c}_{2}}$
For $i=2\Rightarrow
{{c}_{4}}=\frac{1}{4}{{c}_{3}}$
So $y={{c}_{0}}+{{c}_{1}}x+{{c}_{2}}{{x}^{2}}+{{c}_{3}}{{x}^{3}}+{{c}_{4}}{{x}^{4}}+.....$
$y={{c}_{0}}+{{c}_{1}}x+\frac{1}{2}{{c}_{1}}{{x}^{2}}+\frac{1}{3}{{c}_{2}}{{x}^{3}}+\frac{1}{4}{{c}_{3}}{{x}^{4}}+....$
$y={{c}_{0}}+{{c}_{1}}x+\frac{1}{2}{{c}_{1}}{{x}^{2}}+\frac{1}{3}\times
\frac{1}{2}{{c}_{1}}{{x}^{3}}+\frac{1}{4}\times \frac{1}{3}\times
\frac{1}{2}{{c}_{1}}{{x}^{4}}+....$
Thus $y={{c}_{0}}+{{c}_{1}}\left(
x+\frac{1}{2}{{x}^{2}}+\frac{1}{3}\times \frac{1}{2}{{x}^{3}}+\frac{1}{4}\times
\frac{1}{3}\times \frac{1}{2}{{x}^{4}}+.... \right)={{c}_{0}}+{{c}_{1}}\left(
\frac{x}{1!}+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+\frac{{{x}^{4}}}{4!}+.....
\right)$
Therefore $y={{c}_{0}}+{{c}_{1}}{{e}^{x}}$
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