# Thread: Non-homogeneous Equations Initial Values Problem

1. ## 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

Thank You!!

2. Originally Posted by Phyxius117
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.)