# Differential equation initial value prob

1. Sep 23, 2009

### -EquinoX-

1. The problem statement, all variables and given/known data

http://img16.imageshack.us/img16/1008/questionsh.th.jpg [Broken]

2. Relevant equations

3. The attempt at a solution

My question is that should we divide each side by sin(t) first? Because I tried dividing it with sin(t) and then I have u(x) = exp(csc(x)), which is kind of hard to integrate.

Reason why I divide it by sin(t) is to get the following factor

y' + p(x)y = q(x)

and u(x) = exp(integral of p(x) dx)

Last edited by a moderator: May 4, 2017
2. Sep 23, 2009

### rock.freak667

yes divide by sin(t)

$$\int \frac{cost}{sint} dt$$

notice that d/dt(sint)= cost ?

3. Sep 23, 2009

### -EquinoX-

This will be equal to:

$$\int cot(t) dt$$

which is:

$${-(csc(t))}^2$$

and so u(x) is e^(csc(t))

4. Sep 23, 2009

### CharmedQuark

remember the formula for d/dx(ln x)

d/dx(ln x)= x'/x

separate cot into cos/sin.

Last edited: Sep 23, 2009