Calculating Constant C in a Differential Equation with Initial Condition

[roam]

Homework Statement

http://img546.imageshack.us/img546/1860/dequestion.jpg

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

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

So when we substitute in we get

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

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:

\frac{dx}{dt} + \frac{x}{t} = sin \ 3t

using the integrating factor

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

t \frac{dx}{dt} + t \frac{x}{t}= t(sin \ 3t)

\int \frac{d}{dt} tx = \int t \ sin \ 3t

Using integration by parts for the RHS

tx = \frac{1}{9} \ sin \ 3t - \frac{t}{3} \ cos \ 3t + k

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

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.

 

[tiny-tim]

hi roam!

yes, your general solution is fine

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?

you always use radians

always always always! ​

(though you shouldn't need a calculator for cos3π and sin3π )

 

[Ray Vickson]

Homework Statement

http://img546.imageshack.us/img546/1860/dequestion.jpg

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

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

So when we substitute in we get

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

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:

\frac{dx}{dt} + \frac{x}{t} = sin \ 3t

using the integrating factor

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

t \frac{dx}{dt} + t \frac{x}{t}= t(sin \ 3t)

\int \frac{d}{dt} tx = \int t \ sin \ 3t

Using integration by parts for the RHS

tx = \frac{1}{9} \ sin \ 3t - \frac{t}{3} \ cos \ 3t + k

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

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.

Always use radians in calculus. The formulas become quite messy if you use degrees. For example, if we measure the angle x in degrees we have (d/dx) sin(x) = (π/180)*cos(x) and (d/dx) cos(x) = -(π/180)*sin(x).

RGV

 

[roam]

Ah I see! Thanks a lot tiny tim and Ray for your responses, I get it now!