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Old October 22nd, 2009, 02:20 PM
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Default Non-homogeneous Equations Initial Values Problem

Here is the problem:

y−12y+32y=21ex

y(0)=27 y(0)=21 y(0)=25

Y(x) = ??

I used the method of undetermined coefficients and my answer is:

(71/3)+(8/3)e^(4x)+(-1/3)e^(8x)+e^x

But the online-homework wouldn't accept it and says it's incorrect...

Here is what I did:

Yp(x) = (C1) + (C2)e^(4x) + (C3)e^(8x) + e^x
Y'p(x) = 4(C2)e^(4x) + 8(C3)e^(8x) + e^x
Y''p(x) = 8(C2)e^(4x)+64(C3)e^(8x) + e^x

then after plug in and used the given initial values I got:

(C1) + (C2) + (C3) + 1 = 27
4(C2) + 8(C3) + 1 = 21
8(C2) + 64(C3) + 1 = 25

so C1 = 71/3
C2 = 8/3
C3 = -1/3

Please help!!!

Thank You!!
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  #2  
Old October 22nd, 2009, 09:17 PM
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Quote:
Originally Posted by Phyxius117 View Post
Here is the problem:

y−12y+32y=21ex

y(0)=27 y(0)=21 y(0)=25

Y(x) = ??

I used the method of undetermined coefficients and my answer is:

(71/3)+(8/3)e^(4x)+(-1/3)e^(8x)+e^x

But the online-homework wouldn't accept it and says it's incorrect...

Here is what I did:

Yp(x) = (C1) + (C2)e^(4x) + (C3)e^(8x) + e^x
Y'p(x) = 4(C2)e^(4x) + 8(C3)e^(8x) + e^x
Y''p(x) = 8(C2)e^(4x)+64(C3)e^(8x) + e^x (that 8 should be 16)

then after plug in and used the given initial values I got:

(C1) + (C2) + (C3) + 1 = 27
4(C2) + 8(C3) + 1 = 21
8(C2) + 64(C3) + 1 = 25 (again, the 8 should be 16)

so C1 = 71/3
C2 = 8/3
C3 = -1/3 (the coefficients in the complementary solution should be different now.)

Please help!!!

Thank You!!
Note the corrections in red.
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  #3  
Old October 22nd, 2009, 09:38 PM
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Thank you so much!!! Just got the answer correct
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