Solve the given initial-value problem

Question

Answer 1

Let M(x,y) = y²cosx - 3x²y - 2x and N(x,y) = 2ysinx - x³ + lny.

Since ∂M/∂y = ∂N/∂x = 2ycosx - 3x², the equation Mdx + Ndy = 0 is exact

So, there exists a function f(x,y) such that ∂f/∂x = M and ∂f/∂y = N.

Integrate ∂f/∂x = M with respect to x to get f(x,y) = y²sinx - x³y - x² + g(y).

So, ∂f/dy = 2ysinx - x³ + g'(y) = N. Thus, g'(y) = lny.

g(y) = ∫lnydy. Integrate by parts to get g(y) = ylny - y

So, f(x,y) = y²sinx - x³y - x² + ylny - y

Solutions: y²sinx - x³y - x² + ylny - y = c, where c is an arbitrary constant.

Note: the given problem is not an "initial value problem" as no initial value is given!