91Ó°ÊÓ

(a)Given differential equation is,

$x^{3}y\text{'}\text{'}\text{'}+x^{2}y\text{'}\text{'}−2xy\text{'}+2y=0$…(1)

Let,

$\begin{array}{l}y=x^r\\ y\text{'}=r{x}^{r−1}\\ y\text{'}\text{'}=r(r−1)x^{r−2}\\ y\text{'}\text{'}\text{'}=r(r−1)(r−2)x^{r−3}\end{array}$

Substitution in equation (1) we get,

$\begin{array}{l}\left(r{x}^{r−1}\right)+2x^r=0\\ r(r−1)(r−2)x^{r−3+3}+r(r−1)x^{r−2+2}−2r{x}^{r−1+1}+2x^r=0\\ r(r−1)(r−2)x^r+r(r−1)x^r−2r{x}^r+2x^r=0\\ \left(r\right(r−1\left)\right(r−2)+r(r−1)−2r+2)x^r=0\end{array}$

Since $x>0,x^r>0$

$\begin{array}{l}\left(r\right(r−1\left)\right(r−2)+r(r−1)−2r+2)=0\\ r(r−1)(r−2)+r(r−1)−2(r−1)=0\\ (r−1)\left(r\right(r−2)+r−2)=0\\ (r−1)\left(r\right(r−2)+(r−2\left)\right)=0\\ (r−1)(r−2)(r+1)=0\\ ⇒r=1,r=2,r=−1\end{array}$

Therefore,

$y_1=x^1,y_2=x^2,y_3=x^{−1}$

Are all solutions of equation (1)

$\{x^1,x^2,x^{−1}\}$is the fundamental solution set.

(b)Given differential equation is,

$x^4{y}^4+6x^{3}y\text{'}\text{'}\text{'}+2x^{2}y\text{'}\text{'}−4xy\text{'}+4y=0$…(2)

Let,

$\begin{array}{l}y=x^r\\ y\text{'}=r{x}^{r−1}\\ y\text{'}\text{'}=r(r−1)x^{r−2}\\ y\text{'}\text{'}\text{'}=r(r−1)(r−2)x^{r−3}\\ y\text{'}\text{'}\text{'}\text{'}=r(r−1)(r−2)(r−3)x^{r−4}\end{array}$

Substituting in equation (2), we get

$\begin{array}{l}{x}^4\left(r\right(r−1\left)\right(r−2\left)\right(r−3\left)x^{r−4}\right)+6x^3\left(r\right(r−1\left)\right(r−2\left)x^{r−3}\right)+2x^2\left(r\right(r−1\left)x^{r−2}\right)−4x\left(r{x}^{r−1}\right)+4\left(x^r\right)=0\\ r(r−1)(r−2)(r−3)x^{r−4+4}+6r(r−1)(r−3)x^{r−3+3}+2r(r−1)x^{r−2+2}−4r{x}^{r−1+1}+4x^r=0\\ r(r−1)(r−2)(r−3)x^r+6r(r−1)(r−3)x^r+2r(r−1)x^r−4r{x}^r+4x^r=0\\ \left(r\right(r−1\left)\right(r−2\left)\right(r−3)+6r(r−1\left)\right(r−3)+2r(r−1)−4r+4)x^r=0\end{array}$

Since $x>0,x^r>0$

$\begin{array}{l}⇒r(r−1)(r−2)(r−3)+6r(r−1)(r−3)+2r(r−1)−4r+4=0\\ r(r−1)(r−2)(r−3)+6r(r−1)(r−3)+2r(r−1)−4(r−1)=0\\ (r−1)\left(r\right(r−2\left)\right(r−3)+6r(r−2)−2r−4)=0\\ (r−1)\left(r\right(r−2\left)\right(r−3)+6r(r−2)+2(r−2\left)\right)=0\\ (r−1)\left(\right(r−2\left)\right(r(r−3)+6r+2\left)\right)=0\\ (r−1)(r−2)\left(r\right(r−3)+6r+2)=0\\ (r−1)(r−2)(r^2−3r+6r+2)=0\\ (r−1)(r−2)(r^2−3r+2)=0\\ (r−1)(r−2)(r+1)(r+2)=0\\ ⇒r=1,r=2,r=−1,r=−2\end{array}$

Therefore,

$y_1=x^1,y_2=x^2,y_3=x^{−1},y_4=x^{−2}$

Are all solution of equation (2)

$\begin{array}{l}{y}_3=x^{2−3i}\\ =e^{(2−3i)lnx}\\ =x^2\left\{cos\right(−3lnx)+isin(−3lnx\left)\right\}\\ =x^2\left\{cos\right(3lnx)−isin(3lnx\left)\right\}\end{array}$

Hence,

$\{x^1,x^2\{cos\left(3lnx\right)+isin\left(3lnx\right)\},x^2\{cos\left(3lnx\right)−isin\left(3lnx\right)\}$is the fundamental solution set.