Question

Round two:

Answer 1

\[2y''-3y'-2y=1.\]Lambda substitution:\[2\lambda^2-3\lambda-2=0,\]\[(2\lambda+1)(\lambda-2)=0,\]\[\lambda=-\frac{1}{2},\]\[\lambda=2.\]Ergo,\[y=c_{1}e^{-\frac{1}{2}x}+c_{2}e^{2x}.\]RHS is annihilated by\[D(2D^2-3D-2)y=0.\]It follows that\[\lambda(2\lambda^2-3\lambda-2)=0,\]\[\lambda(2\lambda+1)(\lambda-2)=0,\]This has solutions:\[\lambda_{1}=0,\]\[\lambda_{2}=-\frac{1}{2},\]\[\lambda_{3}=2.\]This has the form:\[y=c_{1}+c_{2}e^{-\frac{1}{2}x}+c_{3}e^{2x}.\]But this time,\[y_{p}=A,\]\[y'_{p}=0,\]\[y''_{p}=0.\]Therefore,\[2y_{p}''-3y_{p}'-2y_{p}=-2A=1.\]And\[A=-\frac{1}{2}.\]My final solution should then be:\[y=-\frac{1}{2}+c_{1}e^{-\frac{1}{2}x}+c_{2}e^{2x}.\]Does this seem plausible?

Answer 2

looks correct to me.

Answer 3

The way you found the homogeneous solutions is absolutely right. And your particular solution is also correct, but your method is a bit labored.

Remember that in general:

If p(D) is the formal polynominal of differential operator and the equation you are solving is

p(D)y = e^(at)

then the particular solution is just

yp(t) = e^(at)/p(a)

You'll find a good discussion of this result here:
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-13-finding-particular-sto-inhomogeneous-odes/
****

In your equation, a = 0 and the differential operator is
p(D) = 2D^2 - 3D - 2
hence p(0) = -2

and thus yp(t) = -1/2

Answer 4

Oh! Had I known... -.-

Thanks for telling me that, James!