Show Posts
1
Quiz-3 / TUT0302 QUIZ1
« on: October 11, 2019, 03:22:53 PM »
Q: Verify that the functions y_1 and y_2 are solutions of the given differential equation. Do they constitute a fundamental set of solutions?
y”+4y=0
y_1(t)=cos(2t), y_2(t)=sin(2t)
y_1’(t)=-2sin(2t)
y_1”(t)=-4cos(2t)
y_2’(t)=2cos(2t)
y_2”(t)=-4sin(2t)
Plug y_1 into the given equation
-4cos(2t)+4cos(2t)=0
Plug y_2 into the given equation
-4sin(2t)+4sin(2t)=0
So, y_1 and y_2 are solutions of given equation
W= det(y_1,y_2,y_1',y_2') = y_1(t)y_2’(t) – y_2(t)y_1’(t)
=cos(2t)2cos(2t) – sin(2t)(-2)sin(2t)
=2cos^2(2t)+2sin^2(2t)
2[cos^2(2t)+sin^2(2t)]
=2
Since W≠0, it is a fundamental set of solutions
y”+4y=0
y_1(t)=cos(2t), y_2(t)=sin(2t)
y_1’(t)=-2sin(2t)
y_1”(t)=-4cos(2t)
y_2’(t)=2cos(2t)
y_2”(t)=-4sin(2t)
Plug y_1 into the given equation
-4cos(2t)+4cos(2t)=0
Plug y_2 into the given equation
-4sin(2t)+4sin(2t)=0
So, y_1 and y_2 are solutions of given equation
W= det(y_1,y_2,y_1',y_2') = y_1(t)y_2’(t) – y_2(t)y_1’(t)
=cos(2t)2cos(2t) – sin(2t)(-2)sin(2t)
=2cos^2(2t)+2sin^2(2t)
2[cos^2(2t)+sin^2(2t)]
=2
Since W≠0, it is a fundamental set of solutions