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)|} + CAnd 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 $\ln a -\ln b = \ln \frac a b$.