Suppose y(t) = C1y1(t) + C2y2(t) is the solution of some di?erential equation wi

Question

Suppose y(t) = C1y1(t) + C2y2(t) is the solution of some di?erential equation with
the initial values y(0) = y0 and y'(0) = v0. Find formulas for C1 and C2 in terms of
y1(t), y2(t), y0, v0.

7b) Verify that y3 = 3 cos(2t + p/4) is a solution to the DE, y''+ 4y = 0

7c) We showed in class that y1 = sin(2t) and y2 = cos(2t) forms a fundamental set of
solutions to the di?erential equation in 7b. Use your result from 7a to express the solution
y3 = cos(2t + p/4) in the form y3 = C1 sin(2t) + C2 cos(2t)

Explanation / Answer

A)

y0 = c1y1(0) + c2y2(0) initial values y(0) = y0

v0 = c1y1'(0) + c2y2'(0)

c1 = (y0 y1'(0) - v0 y1(0)) / (y2(0)y1'(0) - y2'(0)y1(0) )
c2 = (y0 y2'(0) - v0 y2(0)) / (y1(0)y2'(0) - y1'(0)y2(0) )

B)

 for verification
  y3 ' = -6sin(2t + pi/4)
 
y3'' = -12cos(2t + pi/4) --------1

4 * y3 = 12 cos(2t + pi/4) ------- 2

adding 1 and 2 we get

y3'' + 4 * y3 =0

hence 3 cos(2t + p/4) is solution of y''+ 4y = 0

 

C)

y1 = sin(2t)  and  y2 = cos(2t)

y1'(0) = -1  and y2'(0) = 0

y0 = 12 and  v0 = -32 

y1(0) = 0  and y2(0) = 1

substitute in above u wil get

therefore c1 = 3/2  and c2 = -3/2  

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