Please show the work so I learn Find a general solution of the system: y1\'=3y1
ID: 1813376 • Letter: P
Question
Please show the work so I learn
Find a general solution of the system:
y1'=3y1 - 4y2 + 20cos(t)
y'2=y(base)1 - 2y(base)2
a.) y1=c1e^-t + 4c2e^t - 14cos(t) + 2sin(t),
y2=c1e^-t + c2e^t - 6cos(t) - 2sin(t)
b.) y1=c1e^-t + 4c2e^2t - 14cos(t) + 2sin(t),
y2=c1e^-t + c2e^2t - 6cos(t) - 2sin(t)
c.) y1=c1e^-t + 4c2e^2t + 14cos(t) + 2sin(t),
y2=c1e^-t + c2e^2t + 6cos(t) - 2sin(t)
d.) y1=c1e^6t + c2e^-6t - 1/4,
y2=c1e^6t - c2e^-6t - t
e.) None of the above; see Problem Work
Explanation / Answer
y1'=3y1 - 4y2 + 20cos(t)
y'2=y1 - 2y2
matrix A =
[3 -4]
[1 -2]
eigen values :
|y-3 4|
|-1 y+2| = 0
==>
(y-3)(y+2) + 4 = 0
==>
y^2 - y - 2 = 0
==>
y = 2, -1
eigen vector of y = -1 is
[-4 4] [x] = [0]
[-1 1] [y] [0]
==>
x = y
eigen vector V1 =
[1]
[1]
for y = 2
[-1 4] [x] = [0]
[-1 4] [y] [0]
===>
x = 4y
Eigen vector is V2 =
[4]
[1]
Complementary solution is
Yc = C1*e^(-t) * V1 + C2 * e^(2t)*V2
==>
Yc = C1*e^-t *
[1]
[1]
+ C2*e^(2t)*
[4]
[1]