Question

Show that the function
y=c_1 e^x cos(x)+ c_2 e^x sin(x) satisfies the differential equation
y''-2y'+2y=0 for any values of c1 and c2, then find the values for those constants that solve the initial value problem y(0)= 1, y'(0)= -1

Answer 1

y=c1 e^x cos(x)+ c2 e^x sin(x)
y(0) = 1 , so
1 = c1 e^0 cos 0 + c2 e^0 sin 0
1 = c1 + 0
c1 = 1

y' = c1 e^x (-sin(x)) + c1 e^x cosx + c2 e^x cos(x) + c2 e^x sinx
= e^x( - c1
y'(0) = -1
-1 = c1 e^0 (-sin 0) + c2 e^0 cos 0
-1 = 0 + c2
c2 = -1

y'' = c1 e^x (-cos(x)) + c2 e^x (-sin(x))

then :
y'' - 2y' + 2y
= e^x(2 c2 cosx – 2 c1 sinx) – 2e^x(c1 cosx + c2 sinx – c1 sinx + c2 cosx) + 2e^x(c1 cosx + c2 sinx)
= e^x( 2 c2cosx – 2 c1sinx – 2 c1cosx - 2c2 sinx + 2c1 sinx – 2c2cosx + 2c1cosx + 2c2sinx)
= e^x(0) = 0

YEAHHH