Help with first order linear equations.

Solve the linear equation ( dy(x))/( dx)+y(x) tan(x) = sin(2 x):
Let mu(x)  =  e^( integral tan(x) dx)  =  sec(x).
Multiply both sides by mu(x):
sec(x) ( dy(x))/( dx)+(sec(x) tan(x)) y(x)  =  sec(x) sin(2 x)
Substitute tan(x) sec(x)  =  ( dsec(x))/( dx):
sec(x) ( dy(x))/( dx)+( dsec(x))/( dx) y(x)  =  sec(x) sin(2 x)
Apply the reverse product rule g ( df)/( dx)+f ( dg)/( dx)  =  ( d)/( dx)(f g) to the left-hand side:
( d)/( dx)(sec(x) y(x))  =  sec(x) sin(2 x)
Integrate both sides with respect to x:
integral ( d)/( dx)(sec(x) y(x)) dx  =   integral sec(x) sin(2 x) dx
Evaluate the integrals:
sec(x) y(x)  =  -2 cos(x)+c_1, where c_1 is an arbitrary constant.
Divide both sides by mu(x)  =  sec(x):
Answer: | y(x)  =  cos(x) (-2 cos(x)+c_1)

Oh right, I see. Thankyou

then if you apply the boundary condition you get;

\(y = -2cos^2(x) + 3cos(x)\)