find the general solution of the differential equation and use it to determine solutions as t - infinity

y'+3y=t+e^(-2t)

Question

find the general solution of the differential equation and use it to determine solutions as t -> infinity

y'+3y=t+e^(-2t)

amistre64 · Answer

what is the homogenous solution?

amistre64 · Answer

y' + 3y = 0

anonymous · Answer

$$e ^{3t}y=\int\limits_{?}^{?}e ^{3t}*(t+e ^{-2t})dt$$ is as far as i got

anonymous · Answer

my integrating factor was $$\mu(t)= e ^{\int\limits(3(t)dt}$$

amistre64 · Answer

$$y_h=A~e^{rt}$$$$r+3=0~:~r=-3$$$$y_h=A~e^{-3t}$$

$$y_p=A(t)~e^{-3t}$$$$y'_p=A'(t)~e^{-3t}-3A(t)e^{-3t}~:~A'(t)~e^{-3t}=0$$$$y'_p=-3A(t)e^{-3t}$$

its been awhile since i tried it using an I.F.

anonymous · Answer

well the answer is $$y=ce ^{-3t}+\frac{ t }{ 3 }-\frac{ 1 }{ 9 }+e ^{-2t}$$

anonymous · Answer

i got it from the back of the book but have no idea how to get there from where i am now

amistre64 · Answer

$$A'(t)e^{-3t}=t+e^{-2t}$$
$$A'(t)=te^{3t}+e^{(-2+3)t}$$
$$\int A'(t)dt=\int te^{3t}+e^{(-2+3)t}dt$$
$$\int A'(t)dt=\frac{t}{3}e^3-\frac{1}{9}e^{3t}+e^{t}+C$$
$$y=c~e^{-3t}+\frac{t}{3}e^3-\frac{1}{9}e^{3t}+e^{t}+C$$

pfft, i can get someplace close to it, but i doubt my method was sound

anonymous · Answer

im still not understanding where u got this A from

amistre64 · Answer

$$y=c~e^{-3t}+(\frac{t}{3}-\frac{1}{9}+e^{-2t})e^{-3t}$$

amistre64 · Answer

A is just something i used to call "c" by

anonymous · Answer

o ok thank you very much

amistre64 · Answer

somehow i either get an extra  $e^{-3t}$, but it might be appropriate ... i just dont see how lol

anonymous · Answer

was my integrating factor and my first equation setup before i integrated right?

amistre64 · Answer

should used e^(-3t), but from what i can recall, it looks like a good setup otherwise

anonymous · Answer

alright but i still dont understand how you do this: $$\int\limits(e ^{3t}(t+e ^{-2t})dt$$

amistre64 · Answer

y = Ce^(-3t)
y' = -3Ce^(-3t)

y' + 3y = 0

-3Ce^(-3t) + 3Ce^(-3t) = 0

amistre64 · Answer

had connection issues with the site

amistre64 · Answer

distribute it thru and integrate the pieces

anonymous · Answer

i got $$e ^{3t}t+e ^{t}$$

amistre64 · Answer

good, not the last term is simple enough, the first term takes it by parts

amistre64 · Answer

*now

amistre64 · Answer

u = t        v = e^3t/3
du = dt    dv = e^3t dt


{S} te^(3t) dt = t e^3t/3 - {S} e^3t/3 dt

{S} te^(3t) dt = t e^3t/3 - e^3t/9

{S} te^(3t) dt = e^3t( t/3 - 1/9)

amistre64 · Answer

i spose +C is in order :)

e^3t( t/3 - 1/9) + e^t + C

now divide off that e^3t from the front

( t/3 - 1/9) + e^-2t + Ce^-3t

anonymous · Answer

thanks for your help but im totally lost, i think ill just go back to clac 2

amistre64 · Answer

what part did you get lost at?  or would this look better in the coding?

anonymous · Answer

well i got 
te^3t+ e^t + C and u got e^3t (t/3-1/9) + C

amistre64 · Answer

$$e^{3t}y=\int te^{3t}dt+\int e^tdt$$
$$e^{3t}y=\int te^{3t}dt+e^t+c$$
$$e^{3t}y=\frac{1}{3}te^{3t}-\frac{1}{3}\int e^{3t}dt+e^t+c$$
$$e^{3t}y=\frac{1}{3}te^{3t}-\frac{1}{3}\frac{1}{3}e^{3t}+c+e^t+c$$
$$e^{3t}y=\frac{t}{3}e^{3t}-\frac{1}{9}e^{3t}+e^t+C$$divide off that front e^3t$$y=\frac{t}{3}\frac{e^{3t}}{e^{3t}}-\frac{1}{9}\frac{e^{3t}}{e^{3t}}+\frac{e^t}{e^{3t}}+\frac{C}{e^{3t}}$$
$$y=\frac{t}{3}-\frac{1}{9}+e^{-2t}+Ce^{-3t}$$

anonymous · Answer

I got what "the back of the book" got, but I don't understand the last part of the question:

$\quad$"Determine solutions as  $t\rightarrow\infty$"

The expression  $y=\frac{1}{3}t-\frac{1}{9}+e^{-2t}+ke^{-3t}$  clearly diverges as $t\rightarrow\infty$

anonymous · Answer

omg yes finally thank you soooooo much amistre64, i may have hope after all

amistre64 · Answer

:)  good luck