anonymous 5 years ago y''+5y'+6y=2e^t

1. anonymous

$yh(t) = \alpha e ^{2t}+\beta e ^{3t}$ $yp(t) = 1/20e ^{2t}$ Therefore $Y(t) = \alpha e ^{2t} + \beta e ^{3t} + 1/20e ^{2t}$ Look right?

2. amistre64

this one?

3. amistre64

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

4. anonymous

That's cool. Thanks for taking a look.

5. anonymous

It does not look right to me.

6. anonymous

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: $y_c=c_1e^{-2t}+c_2e^{-3t}$

7. anonymous

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: $y=y_c+y_p=c_1e^{-2t}+c_2e^{-3t}+{e^t \over 6}$

8. anonymous

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.

9. anonymous