Find the value of k so that the differential equation is exact.

Question

(2xy² + ye^x)dx + (2x²y + ke^x − 1)dy = 0

Answer 1

A differential equation Mdx + Ndy = 0 is exact only when ∂M/∂y = ∂N∂x.

So, for (2xy² + ye^x)dx + (2x²y + ke^x - 1)dy = 0 to be exact we must have:

4xy + e^x = 4xy + ke^x. Therefore, k = 1.