# Homogeneous DE of the first order

• Feb 23rd 2010, 06:35 PM
arbolis
Homogeneous DE of the first order
I must solve $\dot x + x\cos t=0$.
I realize it's separable, and solving it I reach $x=e^{-\sin (t) + C}$ where $C$ is a constant.

However, solving it via the integrating factor method, I do not reach the same result.
The IF is $e^{\sin t}$.
Multiplying both sides by it, I get $\frac{dx}{dt} e^{\sin t}+x\cos (t) e^{\sin t}=0$.
Integrating with respect to t, I get $e^{\sin t}x=0$.
Now clearly $x\equiv 0$.
What am I doing wrong?
• Feb 23rd 2010, 07:20 PM
Chris L T521
Quote:

Originally Posted by arbolis
I must solve $\dot x + x\cos t=0$.
I realize it's separable, and solving it I reach $x=e^{-\sin (t) + C}$ where $C$ is a constant.

However, solving it via the integrating factor method, I do not reach the same result.
The IF is $e^{\sin t}$.
Multiplying both sides by it, I get $\frac{dx}{dt} e^{\sin t}+x\cos (t) e^{\sin t}=0$.
Integrating with respect to t, I get $e^{\sin t}x=0$.
Now clearly $x\equiv 0$.
What am I doing wrong?

What I think we need to do at $\frac{\,d}{\,dt}\left[xe^{\sin t}\right]=0$ is integrate both sides and say that $xe^{\sin t}={\color{red}C}$ in order to avoid getting the trivial solution for an answer. From here then, we get $x=Ce^{-\sin t}$ which is the answer you got from solving it using separation of variables.

Does this help?
• Feb 23rd 2010, 07:23 PM
arbolis
Quote:

Originally Posted by Chris L T521
What I think we need to do at $\frac{\,d}{\,dt}\left[xe^{\sin t}\right]=0$ is integrate both sides and say that $xe^{\sin t}={\color{red}C}$ in order to avoid getting the trivial solution for an answer. From here then, we get $x=Ce^{-\sin t}$ which is the answer you got from solving it using separation of variables.

Does this help?

Bright! (Clapping)
Yes it helps, a lot.
By the way your tutorial is really great! I'm going to use it often these days.