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## Homework Statement

http://img546.imageshack.us/img546/1860/dequestion.jpg [Broken]

## The Attempt at a Solution

My main problem is with the second part where it says find a solution to the DE that satisfies the initial condition x(π) = 2. I found the general solution to be

[itex]x(t) = \frac{1}{9 t} \ sin \ 3t - \frac{1}{3} \ cos \ 3t + \frac{c}{t}[/itex]

So when we substitute in we get

[itex]x(\pi) = 2 = \frac{1}{9 \pi} \ sin \ 3 \pi - \frac{1}{3} \ cos \ 3 \pi + \frac{c}{\pi}[/itex]

Now I need to solve for the constant C. So my question is: Do I need to calculate this using radians or degrees? I mean, should I have my calculators settings on degrees or radians?

By the way this is how I solved the DE:

[itex]\frac{dx}{dt} + \frac{x}{t} = sin \ 3t[/itex]

using the integrating factor

[itex]\mu (t) = e^{\int \frac{1}{t} dt}= k \ e^{\ln |t|} = t[/itex]

[itex]t \frac{dx}{dt} + t \frac{x}{t}= t(sin \ 3t)[/itex]

[itex]\int \frac{d}{dt} tx = \int t \ sin \ 3t[/itex]

Using integration by parts for the RHS

[itex]tx = \frac{1}{9} \ sin \ 3t - \frac{t}{3} \ cos \ 3t + k[/itex]

[itex]\therefore \ x(t) = \frac{1}{9 t} \ sin \ 3t - \frac{1}{3} \ cos \ 3t + \frac{c}{t}[/itex]

Is this correct? I think I solved it correctly, but I would still appreciate it if anyone could let me know if there are any mistakes. I'm a bit unsure because most of the DE solutions I've seen don't have a variable in the denominator like the 1/9t term.

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