How Do You Solve This Tricky Differential Equation Problem?

[cklabyrinth]

Homework Statement

$y'sin(2t) = 2(y+cos(t))$

$y(\frac{∏}{4}) = 0$

Homework Equations

$\frac{dy}{dx} + p(x)y = q(x)$

$y = \frac{\int u(x) q(x) dx + C}{u(x)}$

where
$u(x) = exp(\int p(x)dx)$

The Attempt at a Solution

I've set the equation in the form above, simplified the RHS and solved for u(x):

$\frac{dy}{dt} - \frac{2}{sin(2t)}y = \frac{2cos(t)}{sin(2t)}$RHS simplification:
$\frac{2cos(t)}{sin(2t)} = \frac{2cos(t)}{2cos(t)sin(t)} = csc(t)$

which gives:

$\frac{dy}{dt} - \frac{2}{sin(2t)}y = csc(t)$
$u(x) = exp(\int -\frac{2}{sin(2t)}dt)$

u-sub with u = 2t in the integration gives:

$u(x) = exp(\int -csc(u)du) = exp(ln|cos(2t)|) = |cos(2t)|$
then:

$y = \frac{\int |cos(2t)|csc(t)dt}{|cos(2t)|} + C$And I'm stuck here on the indefinite integral in the numerator:

$\int |cos(2t)|csc(t)dt$

I've tried replacing cos(2t) with $cos^2(t) - sin^2(t)$ but that ends up leaving me with $\int cot(t)cos(t)dt - \int sin(t)dt$ which I find just as hard. Any ideas on where I'm going wrong?

 

[LCKurtz]

Homework Statement

$y'sin(2t) = 2(y+cos(t))$

$y(\frac{∏}{4}) = 0$

Homework Equations

$\frac{dy}{dx} + p(x)y = q(x)$

$y = \frac{\int u(x) q(x) dx + C}{u(x)}$

where
$u(x) = exp(\int p(x)dx)$

The Attempt at a Solution

I've set the equation in the form above, simplified the RHS and solved for u(x):

$\frac{dy}{dt} - \frac{2}{sin(2t)}y = \frac{2cos(t)}{sin(2t)}$


RHS simplification:
$\frac{2cos(t)}{sin(2t)} = \frac{2cos(t)}{2cos(t)sin(t)} = csc(t)$

which gives:

$\frac{dy}{dt} - \frac{2}{sin(2t)}y = csc(t)$



$u(x) = exp(\int -\frac{2}{sin(2t)}dt)$

u-sub with u = 2t in the integration gives:

$u(x) = exp(\int -csc(u)du) = exp(ln|cos(2t)|) = |cos(2t)|$

That isn't the correct antiderivative for the csc integral.

 

[cklabyrinth]

I did a bit of simplification, but I may have done something wrong in between:

$\int -csc(u)du = -ln|csc(u) - cot(u)| = ln|csc(2t) - cot(2t)|$

$=-ln|\frac{csc(2t)}{cot(2t)}| = -ln|\frac{1}{sin(2t)}*\frac{sin(2t)}{cos(2t)}|$

$= -ln|sec(2t)| = ln|sec(2t)|^-1 = ln|cos(2t)|$

When exponentiating that, I get to $u(x) = |cos(2t)|$

Did I do something wrong along the way?

 

[LCKurtz]

I did a bit of simplification, but I may have done something wrong in between:

$\int -csc(u)du = -ln|csc(u) - cot(u)| = ln|csc(2t) - cot(2t)|$

$=-ln|\frac{csc(2t)}{cot(2t)}|$

The correct formula for logarithms is $\ln a -\ln b = \ln \frac a b$.