Download + y = 0, if we compare this to equation

MATH 2930 HW#14 Solution
2/25/2013
In all problems t>0.
35. t2y'' + ty' + y = 0, if we compare this to equation (ii) in problem 34, we see that α=1 and β=1.
d2 y
dy
d2 y
+(α−1)
+β
y=
+ y=0.
If we let x = ln(t), we can obtain
dx
dx 2
dx 2
The two roots of the characteristic equation of this linear second-order differential equation with
±√ −4
=±i. The general solution is then
constant coefficients are r =
2
y=c 1 sin ( x )+c 2 cos ( x)=c 1 sin (ln t)+c 2 cos (ln t). Remember to express the solution in terms of t!
36. t2y'' + 4ty' + 2y = 0, if we compare this to equation (ii) in problem 34, we see that α=4 and β=2.
d2 y
dy
d2 y
dy
+(α−1) +β y= 2 +3 +2y=0.
If we let x = ln(t), we can obtain
2
dx
dx
dx
dx
The two roots of the characteristic equation of this linear second-order differential equation with
−3±√ 9−8
=−1,−2. The general solution is then
constant coefficients are r =
2
c c
y=c 1 e−x +c2 e−2x=c1 e−ln t+c 2 e−2 ln t = 1 + 22 .
t t
37. t2y'' + 3ty' + 1.25y = 0, so we see that α=3 and β=1.25.
d2 y
dy
d2 y
dy
+(α−1)
+β
y=
+2 +1.25y=0.
If we let x = ln(t), we can obtain
2
2
dx
dx
dx
dx
The two roots of the characteristic equation of this linear second-order differential equation with
−2±√ 4−5
1
=−1± i. The general solution is then
constant coefficients are r =
2
2
c1
1
1
1
1
ln t c 2
ln t
−x
−x
−ln t
−ln t
y=c 1 e sin ( x)+c 2 e cos ( x)=c1 e sin( ln t)+c 2 e cos ( ln t )= sin (
)+ cos (
).
2
2
2
2
t
2
t
2