Show that the differential equation 2x sin 3y dy + 3x^2 cos 3y = 0 is exact and hence solve.

Steps to Solve

1. Identify the components of the equation
   The given differential equation is:
   $$ 2x \sin(3y) dy + 3x^2 \cos(3y) = 0 $$
   We can rewrite it as:
   $$ M(x, y) = 3x^2 \cos(3y) \quad \text{and} \quad N(x, y) = 2x \sin(3y) $$

2. Check for exactness condition
   A differential equation is exact if:
   $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$
   Calculate:

• For (M):
  $$ \frac{\partial M}{\partial y} = -9x^2 \sin(3y) $$
• For (N):
  $$ \frac{\partial N}{\partial x} = 2 \sin(3y) $$

Since ( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} ), the equation is not exact.

3. Differentiate with respect to y and x
   Now let's recalculate to check our derivatives correctly:
   For ( M ):
   $$ \frac{\partial M}{\partial y} = 3x^2 \cdot (-3 \sin(3y)) = -9x^2 \sin(3y) $$
   For ( N ):
   $$ \frac{\partial N}{\partial x} = 2 \sin(3y) $$

These derivatives do not equalize, indicating that the equation is indeed not exact.

4. Finding an integrating factor
   We can often find an integrating factor, but it's not straightforward here; however, let's examine if a function of (y) or (x) can work. In some cases, a function related to (x) or (y) van be used.

5. Using potential functions
   Alternatively, than solving directly, we look for potential solutions. This implies finding a function (F(x, y)) such that:
   $$ \frac{\partial F}{\partial x} = M(x, y) \quad \text{and} \quad \frac{\partial F}{\partial y} = N(x, y) $$

6. Integrating (M)
   We can integrate (M(x, y)) with respect to (x):
   $$ F(x, y) = \int 3x^2 \cos(3y) dx = x^3 \cos(3y) + g(y) $$
   Where (g(y)) is an arbitrary function of (y).

7. Differentiating (F) with respect to (y)
   Next, we differentiate (F) with respect to (y):
   $$ \frac{\partial F}{\partial y} = -3x^3 \sin(3y) + g'(y) $$
   Setting this equal to (N(x, y)):
   $$ -3x^3 \sin(3y) + g'(y) = 2x \sin(3y) $$

8. Solving for (g(y))
   We can equate the remaining terms to find (g'(y)):
   $$ g'(y) = 2x \sin(3y) + 3x^3 \sin(3y) $$
   This means that:

9. Final form of solution
   Integrating, we can find an implicit solution.