1. **State the problem:** Solve the differential equation $$(x - y)(4x + y) \, dx + x(5x - y) \, dy = 0.$$ We want to find the general solution and identify which of the given options matches it. 2. **Rewrite the equation:** The differential equation can be written as: $$M(x,y) = (x - y)(4x + y), \quad N(x,y) = x(5x - y)$$ where $M$ is the coefficient of $dx$ and $N$ is the coefficient of $dy$. 3. **Check if the equation is exact:** Compute partial derivatives: $$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}[(x - y)(4x + y)] = (x - y) \cdot 1 + (4x + y)(-1) = (x - y) - (4x + y) = x - y - 4x - y = -3x - 2y,$$ $$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}[x(5x - y)] = (5x - y) + x \cdot 5 = 5x - y + 5x = 10x - y.$$ Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the equation is not exact. 4. **Try to find an integrating factor:** Check if an integrating factor depends on $x$ or $y$ alone. Calculate: $$\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = (10x - y) - (-3x - 2y) = 10x - y + 3x + 2y = 13x + y.$$ Check if $\frac{13x + y}{M} = \frac{13x + y}{(x - y)(4x + y)}$ is a function of $x$ alone or $y$ alone. It depends on both variables, so no simple integrating factor of $x$ or $y$ alone. 5. **Try substitution:** Let $$z = \frac{y}{x} \implies y = zx, \quad dy = z \, dx + x \, dz.$$ Rewrite the equation in terms of $x$ and $z$: Calculate each term: $$(x - y) = x - zx = x(1 - z),$$ $$(4x + y) = 4x + zx = x(4 + z),$$ $$x(5x - y) = x(5x - zx) = x^2(5 - z).$$ Substitute into the original equation: $$M \, dx + N \, dy = (x(1 - z))(x(4 + z)) \, dx + x^2(5 - z)(z \, dx + x \, dz) = 0,$$ which simplifies to: $$x^2(1 - z)(4 + z) \, dx + x^2(5 - z)z \, dx + x^3(5 - z) \, dz = 0,$$ $$x^2[(1 - z)(4 + z) + (5 - z)z] \, dx + x^3(5 - z) \, dz = 0.$$ 6. **Simplify the coefficient of $dx$:** $$(1 - z)(4 + z) + (5 - z)z = (4 + z - 4z - z^2) + (5z - z^2) = (4 - 3z - z^2) + (5z - z^2) = 4 + 2z - 2z^2.$$ So the equation becomes: $$x^2(4 + 2z - 2z^2) \, dx + x^3(5 - z) \, dz = 0.$$ Divide both sides by $x^3$: $$\frac{x^2}{x^3}(4 + 2z - 2z^2) \, dx + (5 - z) \, dz = 0,$$ $$\frac{4 + 2z - 2z^2}{x} \, dx + (5 - z) \, dz = 0.$$ 7. **Rewrite as:** $$(4 + 2z - 2z^2) \, \frac{dx}{x} + (5 - z) \, dz = 0.$$ 8. **Separate variables:** $$\frac{dx}{x} = - \frac{(5 - z)}{4 + 2z - 2z^2} \, dz.$$ 9. **Integrate both sides:** $$\int \frac{dx}{x} = - \int \frac{5 - z}{4 + 2z - 2z^2} \, dz.$$ 10. **Simplify denominator:** $$4 + 2z - 2z^2 = -2z^2 + 2z + 4 = -2(z^2 - z - 2) = -2(z - 2)(z + 1).$$ 11. **Partial fraction decomposition:** $$\frac{5 - z}{4 + 2z - 2z^2} = \frac{5 - z}{-2(z - 2)(z + 1)} = -\frac{5 - z}{2(z - 2)(z + 1)}.$$ Set $$\frac{5 - z}{(z - 2)(z + 1)} = \frac{A}{z - 2} + \frac{B}{z + 1}.$$ Multiply both sides by $(z - 2)(z + 1)$: $$5 - z = A(z + 1) + B(z - 2).$$ 12. **Solve for A and B:** Set $z = 2$: $$5 - 2 = A(2 + 1) + B(0) \implies 3 = 3A \implies A = 1.$$ Set $z = -1$: $$5 - (-1) = A(0) + B(-1 - 2) \implies 6 = -3B \implies B = -2.$$ 13. **Rewrite integral:** $$- \int \frac{5 - z}{2(z - 2)(z + 1)} \, dz = - \frac{1}{2} \int \left( \frac{1}{z - 2} - \frac{2}{z + 1} \right) \, dz = - \frac{1}{2} \left( \int \frac{1}{z - 2} \, dz - 2 \int \frac{1}{z + 1} \, dz \right).$$ 14. **Integrate:** $$- \frac{1}{2} \left( \ln|z - 2| - 2 \ln|z + 1| \right) + C = - \frac{1}{2} \ln|z - 2| + \ln|z + 1| + C.$$ 15. **Recall $z = \frac{y}{x}$:** $$\ln|x| = - \frac{1}{2} \ln \left| \frac{y}{x} - 2 \right| + \ln \left| \frac{y}{x} + 1 \right| + C,$$ $$\ln|x| = \ln \left| \frac{y}{x} + 1 \right| - \frac{1}{2} \ln \left| \frac{y}{x} - 2 \right| + C.$$ 16. **Multiply both sides by 2 to clear fraction:** $$2 \ln|x| = 2 \ln \left| \frac{y}{x} + 1 \right| - \ln \left| \frac{y}{x} - 2 \right| + 2C.$$ 17. **Rewrite logarithms:** $$\ln|x|^2 = \ln \left| \frac{y}{x} + 1 \right|^2 - \ln \left| \frac{y}{x} - 2 \right| + \ln e^{2C}.$$ 18. **Combine logarithms:** $$\ln \left( \frac{|x|^2 \left| \frac{y}{x} + 1 \right|^2}{\left| \frac{y}{x} - 2 \right|} \right) = \ln C',$$ where $C' = e^{2C}$. 19. **Exponentiate both sides:** $$\frac{x^2 \left( \frac{y}{x} + 1 \right)^2}{\left( \frac{y}{x} - 2 \right)} = C',$$ which simplifies to: $$\frac{x^2 \left( \frac{y + x}{x} \right)^2}{\frac{y - 2x}{x}} = C',$$ $$\frac{x^2 \left( \frac{y + x}{x} \right)^2}{\frac{y - 2x}{x}} = x^2 \frac{(y + x)^2}{x^2} \cdot \frac{x}{y - 2x} = \frac{(y + x)^2 x}{y - 2x} = C'.$$ 20. **Rewrite:** $$x (y + x)^2 = C' (y - 2x).$$ 21. **Replace $C'$ by $c$ and rearrange:** $$x (x + y)^2 = c (2x - y).$$ 22. **Match with options:** This matches option C: $$\boxed{x(x + y)^2 = c(2x - y)}.$$