Solve the initial-value problem?

Question

dy/dx = 2/5sin(2x)e^cos x y^2 (y > 0),

y = 1 when x = 0,

giving the solution in explicit form.

Answer 1

I think your equation is dy/dx = (2/5)*(sin (2x))*(e^(cos x))*y^2.

The equation is separable, and can be rewritten as
y^(-2)dy = 2*(2/5)*(sinx)*(cos x)e^(cos x)dx, where we have used the double angle formula for sin (2x). To integrate the right side, substitute w = cos x, and find
-y^(-1) = -(4/5)*(cos x - 1)e^(cos x) + C. Using the initial conditions, find C = -1.

Answer 2

Sorry i dont really have a clue about this, i did do higher maths, but that was a long long time ago.

But if your saying Y is 1 when X is 0 then the first part (dy/dx) basically looks to me to wrong as you cannot divide by zero.
Regardless of what "d" is the start of the question is 1/0

Sorry


Al....