- #1
bobmerhebi
- 38
- 0
1. Solve by Seperable Variable: sin3xdx + 2ycos33xdy = 0
2. We need write it in the form: dy/dx = g(x).h(y)
3. The Attempt at a Solution :
a few algebraic operations lead to: dy/dx = (-sin3x/cos33x).(1/2y) with h(y) = 1/2y
now we have (y2)' = 2ydy = -sin3xdx/cos33x
integrating we get: y2 = [tex]\int[/tex]-sin3xdx/cos33x =G(x)
thus y2 = (1/3).[tex]\int[/tex]d(cos3x)/cos33x = (1/3).(-1/2).(1/cos23x)
so y2 = -1/6cos23x.
here i reached a dead end. is there anything wrong in the attempt or is it right as i think & the given d.e. is wrong ? note that i tried to solve it more then 10 times, i also tried the tan3x & its derivative but the -ve sign persists.
waiting for your help. thanks in advance
2. We need write it in the form: dy/dx = g(x).h(y)
3. The Attempt at a Solution :
a few algebraic operations lead to: dy/dx = (-sin3x/cos33x).(1/2y) with h(y) = 1/2y
now we have (y2)' = 2ydy = -sin3xdx/cos33x
integrating we get: y2 = [tex]\int[/tex]-sin3xdx/cos33x =G(x)
thus y2 = (1/3).[tex]\int[/tex]d(cos3x)/cos33x = (1/3).(-1/2).(1/cos23x)
so y2 = -1/6cos23x.
here i reached a dead end. is there anything wrong in the attempt or is it right as i think & the given d.e. is wrong ? note that i tried to solve it more then 10 times, i also tried the tan3x & its derivative but the -ve sign persists.
waiting for your help. thanks in advance