a) Solve char eq to get fund solns y1 ... y6:
r^6 + 4r^2 = 0 ... ->... r^2 ( r^4 + 4 ) = 0 ... ->...
r^2 = 0 and r^4 = -4.
The six roots are
r = 0 ... 0 ... 1 + i ... 1 - i ... -1 + i ... -1 - i ...
(repeated real and two distinct complex pairs).
The general solution of the ODE is:
y = C1 + C2 t + C3 e^t cos(t) + C4 e^t sin(t)
+ C5 e^(-t) cos(t) + C6 e^(-t) sin(t).
b) First solve char eq to get fund solns y1 ... y2:
r^2 + 6r + 13 = 0.
The two roots are
r = -3 + 2i ... -3 - 2i ... (complex pair).
The two fund solns of the homogeneous ODE are:
y1 = e^(-3t) cos(2t) ... y2 = e^(-3t) sin(2t) .
To find a particular solution make the guess
yp = t^s { (At + B)e^t } + t^z { C } = (At + B)e^t + C
... (i.e. can take s=0, z=0).
Solving for A, B, C gives A = 1/20 ... B = -1/50 ... C = 1/13.
The general solution of the ODE is:
y = C1 e^(-3t) cos(2t) + C2 e^(-3t) sin(2t)
+ [ t/20 - 1/50 ] e^t + 1/13.
c) First solve char eq to get fund solns y1 ... y2:
r^2 + 4r + 4 = 0.
The two roots are
r = -2 ... -2 ... (repeated real).
The two fund solns of the homogeneous ODE are:
y1 = e^(-2t) ... y2 = t e^(-2t) .
Notice that g = e^{-2t} / t^2 isn't a "simple combination"
(i.e. sum or product) of exponentials, polynomials or trig
functions. Rather than try and guess a form for yp we can just
use the variation of parameters method:
yp = - y1 int{g y2 / W } + y2 int{ g y1 / W}
where the Wronskian of y1, y2 is W = e^(-4t). Evaluating the
integrals gives
yp = - e^(-2t) ln(t) - e^(-2t) .
The general solution of the ODE is then:
y = C1 e^(-2t) + C2 t e^(-2t) - e^(-2t) ln(t) - e^(-2t)
y = [C1 - 1] e^(-2t) + C2 t e^(-2t) - e^(-2t) ln(t)
.... by combining terms in e^(-2t)
y = D1 e^(-2t) + C2 t e^(-2t) - e^(-2t) ln(t)
.... where D1 is also an arbitrary constant.
a) Fund solns: y1 = 1 ... y2 = e^( t sqrt{2} ) ... y3 = e^( -t sqrt{2} ).
Part soln: yp = - e^t.
Final solution:
y = (1/2) + (1/4) e^( t sqrt{2} ) + (1/4) e^( -t sqrt{2} ) - e^t.
b) Fund solns: y1 = cos(4t) ... y2 = sin(4t).
Part soln: yp = (1/12) cos(2t).
Final soln: y = -(1/12) cos(4t) + (1/12) cos(2t) .
a) Suitable form ... yp = A t^3 + B t^2 + C t + D t^2 e^t.
b) Suitable form ... yp = [A t + B] e^t cos(t) + [C t + D] e^t sin(t).
Assuming y2 = v y1 gives the ODE
t^2 v'' + 4t v' = 0.
Solving for v gives:
v = -(C1 / 3) t^(-3) + C2.
Choosing C1 = -3 and C2 = 0 gives the indep soln
y2 = t^(-2).