Question

y''+2y'=0 find two power series of the given differential equation about ordinary point x=0.Compare the series of the solution with the solution of differential obtained using the method section 4.3.Try to explain any differences between the two forms of the solution.

Answer 1

Consider the differential equation,

$y^{\prime \prime}-y^{\prime}=0\ Let\ y=\sum_{n=0}^{\infty} c_{n} x^{n},$

$then y^{\prime}=\sum_{n=1}^{\infty} n c_{n} x^{n-1} and y^{\prime \prime}=\sum_{n=2}^{\infty} n(n-1) c_{n} x^{n-2}$

Substituting in the differential equation we get

$\begin{aligned} \sum_{n=2}^{\infty} & n(n-1) c_{n} x^{n-2}-\sum_{n=1}^{\infty} n c_{n} x^{n-1}=0 \\ \Rightarrow & \sum_{n=2}^{\infty} \underbrace{n(n-1) c_{n} x^{n-2}}_{k=n-2}-\sum_{n=1}^{\infty} \underbrace{n c_{n} x^{n-1}}_{k=n-1}=0 \\ \sum_{k=0}^{\infty}(k+2)(k+1) c_{k+2} x^{k}-\sum_{k=0}^{\infty}(k+1) c_{k+1} x^{k}=0 \\ & \sum_{k=0}^{\infty}\left[(k+2)(k+1) c_{k+2}-(k+1) c_{k+1}\right] x^{k}=0 \end{aligned}$  

Compare the coefficients on each side to get,

 $\begin{aligned} (k+2)(k+1) c_{k+2}-(k+1) c_{k+1} &=0, k=0,1,2, \ldots \\ c_{k+2} &=\frac{1}{k+2} c_{k+1}, k=0,1,2, \ldots \end{aligned}$

From the relation $c_{k+2}=\frac{1}{k+2} c_{k+1}, k=0,1,2, \ldots$ get,

 $\begin{aligned} c_{2} &=\frac{1}{0+2} c_{0+1} \\ &=\frac{1}{2} c_{1} \\ c_{3} &=\frac{1}{1+2} c_{1+1} \\ &=\frac{1}{3}\left(\frac{1}{2} c_{1}\right) \\ &=\frac{1}{6} c_{1} \\ c_{4} &=\frac{1}{2+2} c_{2+1} \\ &=\frac{1}{4}\left(\frac{1}{6} c_{1}\right) \\ &=\frac{1}{24} c_{1} \end{aligned}$

Rewrite the solution $y_{2}$ as,

 $\begin{aligned} y_{2} &=x+\frac{1}{2 !} x^{2}+\frac{1}{3 !} x^{3}+\frac{1}{4 !} x^{4}+\cdots \\ &=-1+1+x+\frac{1}{2 !} x^{2}+\frac{1}{3 !} x^{3}+\frac{1}{4 !} x^{4}+\cdots \\ &=-1+\left(1+x+\frac{1}{2 !} x^{2}+\frac{1}{3 !} x^{3}+\frac{1}{4 !} x^{4}+\cdots\right) \\ &=-1+\sum_{n=0}^{\infty} \frac{x^{n}}{n !} \\ &=-1+e^{x} \end{aligned}$

Therefore, the two solutions are $y_{1}=1, y_{2}=-1+e^{x}$

Finally general solution to the given differential equation is,

$y(x)=c_{0} \cdot 1+c_{1} \cdot\left(e^{x}-1\right)$