Question

Solve: y' + 6y = t + e^-4t

Answer 1

I'm not quite sure where to start.

Answer 2

well observe that \(y'+6y\) can be multiplied by \(e^{6t}\) to provide conveniently:$$e^{6t}y'+6e^{6t}y=e^{6t}\frac{dy}{dt}+\frac{d(e^{6t})}{dt}y=\frac{d}{dt}(e^{6t}y)$$
\(e^{6t}\) is known as an integrating factor

Answer 3

so multiplying both sides by \(e^{6t}\) yields:$$\frac{d}{dt}(e^{6t}y)=e^{6t}(t+e^{-4t})\\\frac{d}{dt}(e^{6t}y)=te^{6t}+e^{2t}$$now integrate both sides:$$e^{6t}y=\int(te^{6t}+e^{2t})\ dt=\int te^{6t}\ dt+\int e^{2t}\ dt$$

Answer 4

the latter integral is trivial i.e. \(\displaystyle \int e^{2t}\ dt=\frac12e^{2t}+C\)

Answer 5

the former requires integration by parts -- let \(u=t,dv=e^{6t}\ dt\):$$\int te^{6t}\ dt=\frac16 te^{6t}-\frac16\int e^{6t}\ dt=\frac16 te^{6t}-\frac1{36}e^{6t}+C$$

Answer 6

\[y'+6y=t+e ^{-4t}\]
I.F=\[e ^{\int\limits6dt}=e ^{6t}\]
c.s is\[y~e ^{6t}=\int\limits e ^{6t}\left( t+e ^{-4t} \right)dt+c\]

Answer 7

so far we have:$$e^{6t}y=\frac16 te^{6t}-\frac1{36}e^{6t}+\frac12 e^{2t}+C$$now dividing both sides by \(e^{6t}\) yields our solution:$$y=\frac16t-\frac1{36}+\frac12 e^{-4t}+Ce^{-6t}$$

Answer 8

notice that the integrating factor \(e^{6t}\) gives rise to a term \(Ce^{-6t}\) -- this is called the homogeneous solution to the linear problem. it by itself contributes nothing to the solution since \((Ce^{-6t})'+6(Ce^{-6t})=(Ce^{-6t}e^{6t})'=C'=0\)

Answer 9

hello?

Answer 10

Okay, thank you! Sorry I had to run somewhere for a bit.