Question

1. Use the method of reduction of order to find a second solution of the differential equation
t^2 y double prime + 10t y prime - 10y = 0, t > 0 given the solution y1 = t

A. y2 = t^-1
B. y2 = 1
C.y2 = t
D.y2 = t^-10
E.y2 = t^10
F y2 = t^11
G. y2 = t^-11

Answer 1

Assume $y_2(t)=v(t)y_1(t),$ then

$$y_2'(t)=v'(t)y_1(t)+v(t)y_1'(t)=v'(t)t+v(t)$$

$$y_2''(t)=v''(t)t+v'(t)+v'(t)=v''(t)t+2v'(t)$$

Plugging these into the differential equation gives

$$t^2 (v''t+2v')+10t(v't+v)-10vt=0$$

$$t^3v''+2t^2v'+10t^2v'+10tv-10vt=0$$

Since $t>0$

$$tv''+12v'=0$$

Let $v'=u,$ we have

$$tu'+12u=0$$

By separation of variables

$${u'\over u }=-{12\over t}$$

$$u=C_1t^{-12}$$

Thus

$$v'=C_1t^{-12}$$

$$v=-{1\over 11}C_1t^{-11}+C_2$$

$$v=C_3t^{-11}+C_2$$

$$y_2=C_3t^{-10}+C_2t$$

Now we take $C_3=1, C_2=0$

$$y_2=t^{-10}$$

$D.\ \ y_2=t^{-10}$