In my notes, the prof wrote x' = x + sin(t), x(0) = x(adsbygoogle = window.adsbygoogle || []).push({}); _{0}

He did not go through the solution, only wrote that the solution is

x(t,x_{0}) = x_{0}e^{t}+ e^{t}/2 - 1/2(sin(t) + cos(t))

I verified this is correct by substituting.

So when I try to solve this DE, here's what happens:

[tex] {\frac{dx}{dy} \ - \ x = \ sin(t) [/tex]

integrating factor is e^{-t}

[tex] \int \frac{d}{dt} \ (e^{-t}x) = \int e^{-t} \ sin(t) \ dt => [/tex]

[tex] e^{-t} \ x \ = \ \int e^{-t} \ sin(t) \ dt \ = \ -e^{-t} \ cos(t) \ - \ \int cos(t) \ e^{-t} \ dt \ = \ -e^{-t} \ cos(t) - \left( -e^{-t} \ + \int sin(t) \ e^{-t} \ dt \right) [/tex]

...

[tex] x = -(1/2)cos(t) + (1/2)sin(t) [/tex]

So I'm close, but where did I lose my e^{-t}'s ?

**Physics Forums - The Fusion of Science and Community**

# 1st order non homogenous DE problem solving

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: 1st order non homogenous DE problem solving

Loading...

**Physics Forums - The Fusion of Science and Community**