Answer to Question #297892 in Differential Equations for kirthi

Question

find a power series solution of xy'=y

Answer 1

By Power Series Method, the solution of the differential equation is

$y=c_{0} \sum_{n=0}^{\infty} \frac{\left(\frac{x^{2}}{2}\right)^{n}}{n !}=c_{0} e^{\frac{x^{2}}{2}},$ where $c_{0}$ is any constant.

Let us look at some details.

$\text { Let } y=\sum_{n=0}^{\infty} c_{n} x^{n}$

By taking the derivative term by term,

$y^{\prime}=\sum_{n=1}^{\infty} n c_{n} x^{n-1}$

Now, let us look at the differential equation.

$y^{\prime}=x y$

by substituting the above power series in the equation,

$\Rightarrow \sum_{n=1}^{\infty} n c_{n} x^{n-1}=x \cdot \sum_{n=0}^{\infty} c_{n} x^{n}$  

by pulling the first term from the summation on the left,

$\Rightarrow c_{1}+\sum_{n=2}^{\infty} n c_{n} x^{n-1}=\sum_{n=0}^{\infty} c_{n} x^{n+1}$

by shifting the indices of the summation on the left by 2 ,

$\Rightarrow c_{1}+\sum_{n=0}^{\infty}(n+2) c_{n+2} x^{n+1}=\sum_{n=0}^{\infty} c_{n} x^{n+1}$

By matching coefficients,

$c_{1}=0$ and $(n+2) c_{n+2}=c_{n} \Rightarrow c_{n+2}=\frac{c_{n}}{n+2}$

Let us observe the odd terms.

$\begin{aligned} &c_{3}=\frac{c_{1}}{3}=\frac{0}{3}=0 \\ &c_{5}=\frac{c_{3}}{5}=\frac{0}{5}=0\\ &.\\ &.\\ &.\\ &c_{2 n+1}=0 \end{aligned}$

Let us observe the even terms.

$\begin{aligned} &c_{2}=\frac{c_{0}}{2} \\ &c_{4}=\frac{c_{2}}{4}=\frac{c_{0}}{4 \cdot 2}=\frac{c_{0}}{2^{2} \cdot 2 !} \\ &c_{6}=\frac{c_{4}}{6}=\frac{c_{0}}{6 \cdot 4 \cdot 2}=\frac{c_{0}}{2^{3} \cdot 3 !} &c_{2 n}=\frac{c_{0}}{2^{n} \cdot n !} \end{aligned}$

Hence,

$y=\sum_{n=0}^{\infty} \frac{c_{0}}{2^{n} \cdot n !} x^{2 n}=c_{0} \sum_{n=0}^{\infty} \frac{\frac{x^{2 n}}{2^{n}}}{n !}=c_{0} \sum_{n=0}^{\infty} \frac{\left(\frac{x^{2}}{2}\right)^{n}}{n !}=c_{0} e^{\frac{x^{2}}{2}}$

$\text { (by replacing } x \text { by } \frac{x^{2}}{2} \text { in } e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !} \text { ) }$