Homogeneous DE of the first order

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February 23rd 2010, 07:35 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?
February 23rd 2010, 08:20 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?
February 23rd 2010, 08:23 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.