Question

find the intgerating factor fot first linear differential equation
dy/dt +(1+t)*exp(-t)*y = 0

Answer 1

its \(exp [ \int 1+t).e^{-t} dt]\)

but you can separate this too

Answer 2

\[y'+(1+t)e^{-t}y=0\] notice it's in the form \[y'+a(x)y=b(x)\] where b(x) = 0 so the integrating factor for linear equations is \[\huge p(x) = e^{\int\limits a(x) dx}\]

Answer 3

this my solution
\[ e ^{-2e ^{-t}-te ^{-t}} \]

is this the last solution or is there still any step that can be done

Answer 4

Looks good

Answer 5

That is just your integrating factor, you have not however found the general solution for your differential equation

Answer 6

Better do a quick review of "integrating factors," including where they come from and how they are used in solving differential equations.

Answer 7

the general solution of this ODE:
\[\Large u' + P\left( t \right)u = Q\left( t \right)\]
is the subsequent function:
\[\large \begin{gathered}
u\left( t \right) = \exp \left( {\int { - P\left( t \right)dt} } \right)\left\{ {k + \int {Q\left( t \right) \cdot \exp \left( {\int {P\left( t \right)dt} } \right)} } \right\} \hfill \\
\hfill \\
k \in \mathbb{R} \hfill \\
\end{gathered} \]
please see my tutorial:
http://openstudy.com/study#/updates/552b4f50e4b04e5707c1d380
now, in our case, we have:
\[\Large Q\left( t \right) \equiv 0\]
so, we get, as solution:
\[\Large \begin{gathered}
u\left( t \right) = k\exp \left( {\int { - P\left( t \right)dt} } \right) \hfill \\
k \in \mathbb{R} \hfill \\
\end{gathered} \]
furthermore, a simple integration gives the subsequent primitive function:
\[\Large \int { - P\left( t \right)dt} = {e^{ - t}}\left( {t + 2} \right)\]