Question

y''+5y'+6y=2e^t

Answer 1

\[yh(t) = \alpha e ^{2t}+\beta e ^{3t}\]
\[yp(t) = 1/20e ^{2t}\]

Therefore
\[Y(t) = \alpha e ^{2t} + \beta e ^{3t} + 1/20e ^{2t}\]

Look right?

Answer 2

this one?

Answer 3

does it look righht? yes.... can i prove its right...fraid not :)

Answer 4

That's cool. Thanks for taking a look.

Answer 5

It does not look right to me.

Answer 6

The solution of the associated homogeneous equation:
\[y''+5y'+6y=0\] can be found using the auxiliary equation:
\[m^2+5m+6=0 \implies (m+3)(m+2)=0 \implies m=-3, m=-2\]
Hence the complementary solution is:
\[y_c=c_1e^{-2t}+c_2e^{-3t}\]

Answer 7

Let's assume a solution of the form Ae^x for the particular solution, then substituting in the original function gives:
\[Ae^t+5Ae^t+6Ae^t=2e^t \implies 12A=2 \implies A={1 \over 6}\]
Therefore, the particular solution is y_p=e^t/6
That's the general solution of the differential equation is:
\[y=y_c+y_p=c_1e^{-2t}+c_2e^{-3t}+{e^t \over 6}\]

Answer 8

Ok. Thank you. I think I combined two questions into one in my work, and I always misplace those pesky +/- signs. But, my logic seems correct...just my arithmetic & algebra I get lazy and sloppy with. Thank you very much.

Answer 9

I can tell that by looking at your answer. You're welcome.