Question

Differential Equation:

Answer 1

\[y'+4y=3\sin(e^{4x})\]

Answer 2

\[I(x) = e^{\int\limits 4dx} = e^{4x}\]
\[\int\limits [e^{4x}y]' = \int\limits 3e^{4x}\sin(e^{4x})dx\]

Answer 3

u = 4x
du = 4 dx

\[\frac{ 3 }{ 4 } \int\limits e^u \sin(e^u)du\]

Answer 4

And now im stuck :/

Answer 5

What is the derivative of cos(e^u) with respect to u?

Answer 6

\[
\frac
{d}{du}\cos(e^u)= -e^u \sin(e^u)
\]
You should be able to continue now

Answer 7

Oh is it the chain rule?
The derivative of the thing times the inside?\[\cos(e^u)\times(e^u)\]

Answer 8

oops i meant -sin... lol

Answer 9

Your last integral is easy now

Answer 10

Yea got it. Thanks!

Answer 11

put \[e^{4x}=u,e^{4x}4 ~dx=du,\]
\[\int\limits 3 e^{4x} \sin (e^{4x})dx=\frac{ 3 }{ 4 }\int\limits \sin u~du=-\frac{ 3 }{ 4 }\cos u\]