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March 1, 2015

Posted by **Elyse** on Tuesday, July 2, 2013 at 9:08pm.

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

- Calculus-HELP!!!! -
**Reiny**, Tuesday, July 2, 2013 at 10:51pmy=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

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