1. The problem is to define an exact differential equation and provide an example. 2. An exact differential equation is a first-order differential equation of the form $$M(x,y) + N(x,y)\frac{dy}{dx} = 0$$ where there exists a function $$\psi(x,y)$$ such that $$\frac{\partial \psi}{\partial x} = M(x,y)$$ and $$\frac{\partial \psi}{\partial y} = N(x,y)$$. 3. This means the differential equation can be written as $$\frac{d}{dx} \psi(x,y) = 0$$ implying $$\psi(x,y) = C$$ where $$C$$ is a constant. 4. To check if a differential equation is exact, verify if $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$. 5. Example: Consider the differential equation $$ (2xy + 3) + (x^2 + 4y)\frac{dy}{dx} = 0 $$. 6. Here, $$M(x,y) = 2xy + 3$$ and $$N(x,y) = x^2 + 4y$$. 7. Compute partial derivatives: $$\frac{\partial M}{\partial y} = 2x$$ $$\frac{\partial N}{\partial x} = 2x$$ Since they are equal, the equation is exact. 8. Find $$\psi(x,y)$$ by integrating $$M$$ with respect to $$x$$: $$\psi(x,y) = \int (2xy + 3) dx = x^2 y + 3x + h(y)$$ where $$h(y)$$ is a function of $$y$$. 9. Differentiate $$\psi$$ with respect to $$y$$: $$\frac{\partial \psi}{\partial y} = x^2 + h'(y)$$. 10. Set equal to $$N(x,y)$$: $$x^2 + h'(y) = x^2 + 4y \implies h'(y) = 4y$$. 11. Integrate $$h'(y)$$: $$h(y) = 2y^2 + C$$. 12. Therefore, the implicit solution is: $$\psi(x,y) = x^2 y + 3x + 2y^2 = C$$. Final answer: The exact differential equation is $$ (2xy + 3) + (x^2 + 4y)\frac{dy}{dx} = 0 $$ with implicit solution $$ x^2 y + 3x + 2y^2 = C $$.