Question

Question: In Problems 1–10, determine all the singular points of the given differential equation.

5. (t² - t -2)x" + (t +1)x' - (t - 2)x = 0

Short answer

The singular point exists in this differential equation for both ^P(x)and ^Q(x) is at x =2,-1.

Step-by-step solution

Ordinary and Singular Points.

A point ^$x_0$ is called an ordinary point of equation y" +p(x)y' +q(x)y = 0 if both pand qare analytic at $x_0$. If $x_0$is not an ordinary point, it is called a singular point of the equation.

Find the singular points.

The given differential equation is,

(t² - t - 2)x"+(t + t)x' - (t - 2)x = 0

Dividing the above equation by we get,

x"+$\frac{(t+1)}{\text{'}(t^2-t-2)}x\text{'}-\frac{(t-2)}{(t^2-t-2)}x=0$

On comparing the above equation with y"+p(x)y'+q(x)y=0 , we find that,

P(x) =$\frac{(t+1)}{(t^2-t-2)}$

=$\frac{(t+1)}{(t+1)(t-2)}$

=$\frac{1}{(t-2)}$

Q(x) =$-\frac{(t-2)}{(t^2-t-2)}$

=$-\frac{(t-2)}{(t+1)(t-1)}$

=$-\frac{1}{(t+1)}$

Hence, ^P(x)and ^Q(x)are analytic except, perhaps, when their denominators are zero.

For ^P(x)this occurs at x=2 and for Q(x) this occurs at x = -1 .

We see that ^P(x) is actually analytic at ^{x = 0} .

Therefore, ^P(x) is analytic except at x = -1 as well as ^Q(x)is analytic except at x =2.

The singular point exists in this differential equation for both ^P(x)and ^Q(x)is at x = 2,-1.