1. **State the problem:** Solve the differential equation $$(x - y)(4x + y)\,dx + x(5x - y)\,dy = 0.$$\n\n2. **Rewrite the equation:** The given equation is $$(x - y)(4x + y)\,dx + x(5x - y)\,dy = 0.$$\nWe want to find the general solution $y(x)$.\n\n3. **Check if the equation is exact:** Let $$M = (x - y)(4x + y)$$ and $$N = x(5x - y).$$\nCalculate partial derivatives:\n$$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}[(x - y)(4x + y)] = (x - y)\cdot 1 + (4x + y)(-1) = (x - y) - (4x + y) = x - y - 4x - y = -3x - 2y,$$\n$$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}[x(5x - y)] = (5x - y) + x \cdot 5 = 5x - y + 5x = 10x - y.$$\nSince $$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x},$$ the equation is not exact.\n\n4. **Try substitution:** Let $$z = \frac{y}{x} \Rightarrow y = zx.$$ Then, $$dy = z\,dx + x\,dz.$$\nSubstitute into the equation:\n$$(x - y)(4x + y)\,dx + x(5x - y)(z\,dx + x\,dz) = 0.$$\nRewrite terms:\n$$(x - zx)(4x + zx)\,dx + x(5x - zx)(z\,dx + x\,dz) = 0,$$\nSimplify:\n$$(x(1 - z))(x(4 + z))\,dx + x(5x - zx)(z\,dx + x\,dz) = 0,$$\n$$x^2(1 - z)(4 + z)\,dx + x(5x - zx)(z\,dx + x\,dz) = 0,$$\n$$x^2(1 - z)(4 + z)\,dx + x(5 - z)x(z\,dx + x\,dz) = 0,$$\n$$x^2(1 - z)(4 + z)\,dx + x^2(5 - z)z\,dx + x^3(5 - z)\,dz = 0,$$\nGroup $$dx$$ terms:\n$$x^2[(1 - z)(4 + z) + (5 - z)z]\,dx + x^3(5 - z)\,dz = 0.$$\nCalculate inside brackets:\n$$(1 - z)(4 + z) = 4 + z - 4z - z^2 = 4 - 3z - z^2,$$\n$$(5 - z)z = 5z - z^2,$$\nSum:\n$$4 - 3z - z^2 + 5z - z^2 = 4 + 2z - 2z^2.$$\nSo equation becomes:\n$$x^2(4 + 2z - 2z^2)\,dx + x^3(5 - z)\,dz = 0.$$\nDivide both sides by $$x^3$$ (show cancellation):\n$$\frac{\cancel{x^2}}{\cancel{x^3}}(4 + 2z - 2z^2)\,dx + (5 - z)\,dz = 0 \Rightarrow \frac{4 + 2z - 2z^2}{x}\,dx + (5 - z)\,dz = 0.$$\nMultiply both sides by $$x$$ to isolate $$dx$$:\n$$(4 + 2z - 2z^2)\,dx + x(5 - z)\,dz = 0.$$\nRewrite as:\n$$(4 + 2z - 2z^2)\,dx = -x(5 - z)\,dz.$$\nDivide both sides by $$x(4 + 2z - 2z^2)$$ (show cancellation):\n$$\frac{\cancel{(4 + 2z - 2z^2)}}{\cancel{(4 + 2z - 2z^2)}} \frac{dx}{x} = - \frac{5 - z}{4 + 2z - 2z^2} dz.$$\nSo:\n$$\frac{dx}{x} = - \frac{5 - z}{4 + 2z - 2z^2} dz.$$\n\n5. **Integrate both sides:**\n$$\int \frac{dx}{x} = - \int \frac{5 - z}{4 + 2z - 2z^2} dz.$$\nLeft integral is $$\ln|x| + C_1.$$\nSimplify denominator on right:\n$$4 + 2z - 2z^2 = 2(2 + z - z^2) = 2(-(z^2 - z - 2)) = 2(-(z - 2)(z + 1)) = -2(z - 2)(z + 1).$$\nRewrite integral:\n$$- \int \frac{5 - z}{4 + 2z - 2z^2} dz = - \int \frac{5 - z}{-2(z - 2)(z + 1)} dz = \frac{1}{2} \int \frac{5 - z}{(z - 2)(z + 1)} dz.$$\n\n6. **Partial fraction decomposition:**\nSet:\n$$\frac{5 - z}{(z - 2)(z + 1)} = \frac{A}{z - 2} + \frac{B}{z + 1}.$$\nMultiply both sides by $$(z - 2)(z + 1)$$:\n$$5 - z = A(z + 1) + B(z - 2).$$\nExpand:\n$$5 - z = A z + A + B z - 2B = (A + B) z + (A - 2B).$$\nEquate coefficients:\nFor $$z$$: $$-1 = A + B,$$\nFor constant: $$5 = A - 2B.$$\nFrom first: $$B = -1 - A.$$\nSubstitute into second:\n$$5 = A - 2(-1 - A) = A + 2 + 2A = 3A + 2,$$\n$$3A = 3 \Rightarrow A = 1,$$\n$$B = -1 - 1 = -2.$$\n\n7. **Integrate:**\n$$\frac{1}{2} \int \left( \frac{1}{z - 2} - \frac{2}{z + 1} \right) dz = \frac{1}{2} \left( \ln|z - 2| - 2 \ln|z + 1| \right) + C_2.$$\n\n8. **Combine results:**\n$$\ln|x| = \frac{1}{2} \ln|z - 2| - \ln|z + 1| + C,$$\nwhere $$C = C_2 - C_1.$$\nRewrite:\n$$\ln|x| + \ln|z + 1| = \frac{1}{2} \ln|z - 2| + C,$$\n$$\ln|x(z + 1)| = \frac{1}{2} \ln|z - 2| + C,$$\nExponentiate both sides:\n$$|x(z + 1)| = e^C |z - 2|^{1/2}.$$\nLet $$c = e^C > 0,$$\n$$x(z + 1) = c \sqrt{|z - 2|}.$$\nRecall $$z = \frac{y}{x},$$ so:\n$$x \left( \frac{y}{x} + 1 \right) = c \sqrt{\left| \frac{y}{x} - 2 \right|},$$\n$$x \frac{y + x}{x} = c \sqrt{\left| \frac{y - 2x}{x} \right|},$$\n$$y + x = c \sqrt{\frac{|y - 2x|}{x}}.$$\nMultiply both sides by $$\sqrt{x}$$ (assuming $$x > 0$$):\n$$(y + x) \sqrt{x} = c \sqrt{|y - 2x|}.$$\nSquare both sides:\n$$x (y + x)^2 = c^2 (y - 2x).$$\nRewrite as:\n$$x (x + y)^2 = C (2x - y)^2,$$\nwhere $$C = c^2$$ is an arbitrary constant.\n\n9. **Match with options:** This matches option **A:** $$x(x + y)^2 = c(2x - y)^2.$$\n\n**Final answer:** \n$$\boxed{x(x + y)^2 = c(2x - y)^2}.$$