1. **Form the differential equation from** $y = x + \frac{A}{x}$. Step 1: Differentiate $y$ w.r.t. $x$. $$\frac{dy}{dx} = 1 - \frac{A}{x^2}$$ Step 2: Express $A$ from the original function: $$A = x(y - x)$$ Step 3: Substitute $A$ into $\frac{dy}{dx}$: $$\frac{dy}{dx} = 1 - \frac{x(y - x)}{x^2} = 1 - \frac{y - x}{x} = 1 - \frac{y}{x} + 1 = 2 - \frac{y}{x}$$ Step 4: Rearranged forms a differential equation: $$\frac{dy}{dx} + \frac{y}{x} = 2$$ 2. **Find particular solution for** $\frac{dy}{dx} = 3 e^{2x - 3y}$ with $y(0) = 0$. Step 1: Rewrite as separation of variables: $$\frac{dy}{dx} = 3 e^{2x} e^{-3y} \Rightarrow e^{3y} dy = 3 e^{2x} dx$$ Step 2: Integrate both sides: $$\int e^{3y} dy = \int 3 e^{2x} dx$$ Step 3: Integrate: $$\frac{1}{3} e^{3y} = \frac{3}{2} e^{2x} + C$$ Step 4: Apply initial condition $y=0$ when $x=0$: $$\frac{1}{3} e^{0} = \frac{3}{2} e^{0} + C \Rightarrow \frac{1}{3} = \frac{3}{2} + C \Rightarrow C = \frac{1}{3} - \frac{3}{2} = -\frac{7}{6}$$ Step 5: Write implicit solution: $$\frac{1}{3} e^{3y} = \frac{3}{2} e^{2x} - \frac{7}{6}$$ Or explicitly: $$e^{3y} = \frac{9}{2} e^{2x} - \frac{7}{2}$$ 3. **Solve** $\frac{dy}{dx} = \frac{2x}{y+1}$ by separation of variables. Step 1: Rearrange: $$(y+1) dy = 2x dx$$ Step 2: Integrate both sides: $$\int (y+1) dy = \int 2x dx$$ Step 3: Integrate: $$\frac{y^2}{2} + y = x^2 + C$$ Step 4: Multiply by 2 for simplicity: $$y^2 + 2y = 2x^2 + K$$ where $K=2C$ is arbitrary constant. 4. **Show exactness and solve:** $$ (5x^4 + 3x^2 y^2 - 2x y^3) dx + (2x^3 y - 3x^2 y^2 - 5 y^4) dy = 0$$ Step 1: Let $$M = 5x^4 + 3x^2 y^2 - 2x y^3, \quad N = 2x^3 y - 3x^2 y^2 - 5 y^4$$ Step 2: Compute partial derivatives: $$\frac{\partial M}{\partial y} = 6x^2 y - 6x y^2$$ $$\frac{\partial N}{\partial x} = 6x^2 y - 6x y^2$$ Step 3: Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact. Step 4: Find potential function $\psi(x,y)$ where $$\frac{\partial \psi}{\partial x} = M$$ Integrate with respect to $x$: $$\psi = \int (5x^4 + 3x^2 y^2 - 2x y^3) dx = x^5 + x^3 y^2 - x^2 y^3 + h(y)$$ Step 5: Differentiate $\psi$ w.r.t. $y$: $$\frac{\partial \psi}{\partial y} = 2 x^3 y - 3 x^2 y^2 + h'(y)$$ Set equal to $N$: $$2 x^3 y - 3 x^2 y^2 + h'(y) = 2 x^3 y - 3 x^2 y^2 - 5 y^4$$ Step 6: Solve for $h'(y)$: $$h'(y) = -5 y^4$$ Step 7: Integrate: $$h(y) = - y^5 + C$$ Step 8: Write implicit general solution: $$x^5 + x^3 y^2 - x^2 y^3 - y^5 = C$$ 5. **Solve** $\frac{dy}{dx} = \frac{-2x + 5y}{2x + y}$ using substitution $y = vx$. Step 1: Substitute $y = v x$, then $$\frac{dy}{dx} = v + x \frac{dv}{dx}$$ Step 2: Replace in given equation: $$v + x \frac{dv}{dx} = \frac{-2x + 5 v x}{2x + v x} = \frac{x(-2 + 5 v)}{x(2 + v)} = \frac{-2 + 5 v}{2 + v}$$ Step 3: Rearrange to isolate $\frac{dv}{dx}$: $$x \frac{dv}{dx} = \frac{-2 + 5v}{2 + v} - v = \frac{-2 + 5v - v(2 + v)}{2 + v} = \frac{-2 + 5v - 2v - v^2}{2 + v} = \frac{-2 + 3 v - v^2}{2 + v}$$ Step 4: Write as separable differential equation: $$\frac{2 + v}{-2 + 3 v - v^2} dv = \frac{dx}{x}$$ Step 5: Integrate both sides: $$\int \frac{2 + v}{-v^2 + 3 v - 2} dv = \int \frac{dx}{x}$$ Step 6: Factor denominator: $$-v^2 + 3 v - 2 = -(v - 1)(v - 2)$$ Step 7: Use partial fractions and integrate the left side (details omitted for brevity), yielding $$\ln\left| \frac{v - 2}{v - 1} \right| = \ln |x| + C$$ Step 8: Exponentiate both sides: $$\frac{v - 2}{v - 1} = K x$$ Step 9: Back-substitute $v = \frac{y}{x}$: $$\frac{\frac{y}{x} - 2}{\frac{y}{x} - 1} = K x \Rightarrow \frac{y - 2 x}{y - x} = K x$$ This implicit relation represents the general solution. **Answers:** 1. $\frac{dy}{dx} + \frac{y}{x} = 2$ 2. $e^{3y} = \frac{9}{2} e^{2x} - \frac{7}{2}$ 3. $y^2 + 2y = 2x^2 + K$ 4. $x^5 + x^3 y^2 - x^2 y^3 - y^5 = C$ 5. $\frac{y - 2x}{y - x} = K x$