Question

The solution of the differential equation cosx siny dx + sinx cosy dy = 0 is:

• A. sinx - siny = C, where C is a constant.
• B. sinx siny = C, where C is a constant.
• C. cosx cosy = C, where C is a constant.
• D. sinx + siny = C, where C is a constant.

Answer 1

The Correct Option is B

To solve the differential equation cosx siny dx + sinx cosy dy = 0, we can use the method of exact equations. An equation of the form M(x,y)dx + N(x,y)dy = 0 is exact if there exists a function ψ(x,y) such that:

∂ψ/∂x = M and ∂ψ/∂y = N.

Given:

M(x,y) = cosx siny

N(x,y) = sinx cosy

We check if the equation is exact by verifying:

∂M/∂y = ∂(cosx siny)/∂y = cosx cosy

∂N/∂x = ∂(sinx cosy)/∂x = cosx cosy

Since ∂M/∂y = ∂N/∂x, the equation is exact.

Now, we find the function ψ(x,y):

∫M(x,y)dx = ∫cosx siny dx = siny ∫cosx dx = siny sinx + h(y)

Next, we differentiate ψ(x,y) with respect to y:

∂ψ/∂y = sinx cosy + h'(y)

Equating it to N(x,y), we get:

sinx cosy = sinx cosy + h'(y)

This implies h'(y) = 0 which means h(y) is a constant, denoted as C.

Hence, the solution is ψ(x,y) = sinx siny = C.

The correct solution is: sinx siny = C, where C is a constant.