Thread: second order inhomogeneous equations with constant coeffiecients

1. second order inhomogeneous equations with constant coeffiecients

need help with this question plz

2. Originally Posted by mark18
need help with this question plz
This is routine application of the theory that will be in your class notes and/or textbook. Please show your working and where you're stuck.

3. y'' - y' - 2y = 0

let y = e^(kt)
y' = k.e^(kt)
y'' = k^2.e^(kt)

k^2.e^(kt) - k.e^(kt) - 2e^(kt) = 0

e^(kt) (k^2 - k -2) = 0
e^(kt) cant = 0
therefore k^2 - k -2 = 0
k = 2 or -1

therfore Yh = Ae^2t + Be^-t

f(t) = 6t^2 - 1
guess Yp = Ct^2 + Dt + E
Yp' = 2Ct + D
Yp'' = 2C

sub back into original equation
(2C) - (2Ct + D) - 2(Ct^2 + Dt + E) = 6t^2 - 1

2C(1-t-t^2) + D(-1-2t) - 2E = 6t^2 - 1

dont know how to find C D or E

4. Originally Posted by mark18
y'' - y' - 2y = 0

let y = e^(kt)
y' = k.e^(kt)
y'' = k^2.e^(kt)

k^2.e^(kt) - k.e^(kt) - 2e^(kt) = 0

e^(kt) (k^2 - k -2) = 0
e^(kt) cant = 0
therefore k^2 - k -2 = 0
k = 2 or -1

therfore Yh = Ae^2t + Be^-t

f(t) = 6t^2 - 1
guess Yp = Ct^2 + Dt + E
Yp' = 2Ct + D
Yp'' = 2C

sub back into original equation
(2C) - (2Ct + D) - 2(Ct^2 + Dt + E) = 6t^2 - 1

2C(1-t-t^2) + D(-1-2t) - 2E = 6t^2 - 1 Mr F says: Wrong move. The right move is below.

dont know how to find C D or E
$\displaystyle t^2 (-2C) + t(-2C - 2D) + (2C - D - E) = 6t^2 - 1$.

Equate coefficients of powers of t:

-2C = 6 .... (1)

-2C - 2D = 0 .... (2)

2C - D - E = -1 ... (3)

Solve equations (1), (2) and (3) simultaneously.

5. can u tell me if this is correct plz

C = -3
D = 3
E = -8

6. Originally Posted by mark18
can u tell me if this is correct plz

C = -3
D = 3
E = -8
You've solved the three equations correctly.

By the way, I doubt I'm very much better than you at checking by simple substitution into the equations ....