Exact Differential 4Eb111

1. **State the problem:** We are given the differential equation $$\frac{dy}{dx} = \frac{-8e^x \cos(y) + \frac{y^2}{x}}{-8e^x \sin(y) - 2y \ln(x) + 3y^2}$$ and asked to find the implicit solution of the form $$F(x,y) = c$$. 2. **Rewrite the equation:** Multiply both sides by the denominator to get $$\left(-8e^x \sin(y) - 2y \ln(x) + 3y^2\right) dy = \left(-8e^x \cos(y) + \frac{y^2}{x}\right) dx$$ or equivalently $$\left(-8e^x \sin(y) - 2y \ln(x) + 3y^2\right) dy - \left(-8e^x \cos(y) + \frac{y^2}{x}\right) dx = 0$$ 3. **Identify M and N:** Let $$M(x,y) = -\left(-8e^x \cos(y) + \frac{y^2}{x}\right) = 8e^x \cos(y) - \frac{y^2}{x}$$ $$N(x,y) = -8e^x \sin(y) - 2y \ln(x) + 3y^2$$ The differential equation is in the form $$M(x,y) dx + N(x,y) dy = 0$$ 4. **Check if the equation is exact:** Compute partial derivatives $$\frac{\partial M}{\partial y} = -8e^x \sin(y) - \frac{2y}{x}$$ $$\frac{\partial N}{\partial x} = -8e^x \sin(y) - \frac{2y}{x}$$ Since $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$ the equation is exact. 5. **Find the potential function F(x,y):** Integrate $$M(x,y)$$ with respect to $$x$$: $$\int M(x,y) dx = \int \left(8e^x \cos(y) - \frac{y^2}{x}\right) dx = 8 \cos(y) e^x - y^2 \ln(x) + h(y)$$ where $$h(y)$$ is an arbitrary function of $$y$$. 6. **Differentiate F(x,y) with respect to y and equate to N(x,y):** $$\frac{\partial F}{\partial y} = -8e^x \sin(y) - 2y \ln(x) + h'(y) = N(x,y) = -8e^x \sin(y) - 2y \ln(x) + 3y^2$$ This implies $$h'(y) = 3y^2$$ 7. **Integrate to find h(y):** $$h(y) = \int 3y^2 dy = y^3 + C$$ 8. **Write the implicit solution:** $$F(x,y) = 8 e^x \cos(y) - y^2 \ln(x) + y^3 = c$$ where $$c$$ is an arbitrary constant. **Final answer:** $$\boxed{8 e^x \cos(y) - y^2 \ln(x) + y^3 = c}$$