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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[yh(t) = \alpha e ^{2t}+\beta e ^{3t}\] \[yp(t) = 1/20e ^{2t}\] Therefore \[Y(t) = \alpha e ^{2t} + \beta e ^{3t} + 1/20e ^{2t}\] Look right?
this one?
does it look righht? yes.... can i prove its right...fraid not :)

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That's cool. Thanks for taking a look.
It does not look right to me.
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: \[y_c=c_1e^{-2t}+c_2e^{-3t}\]
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: \[y=y_c+y_p=c_1e^{-2t}+c_2e^{-3t}+{e^t \over 6}\]
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.
I can tell that by looking at your answer. You're welcome.

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