I. Variable Separable Differential Equations 1. Solve $y' = x^{-2}$ passing through $(3,2)$. Step 1: Write $y' = \frac{dy}{dx} = x^{-2}$. Separate variables if needed (already separated). Step 2: Integrate both sides: $$y = \int x^{-2} dx = \int x^{-2} dx = -x^{-1} + C = -\frac{1}{x} + C$$ Step 3: Use initial condition $y(3) = 2$: $$2 = -\frac{1}{3} + C \Rightarrow C = 2 + \frac{1}{3} = \frac{7}{3}$$ Step 4: Final solution: $$y = -\frac{1}{x} + \frac{7}{3}$$ 2. Solve $y' = 2xy + 3x - 12y - 18$ with $y(1) = 3$. Step 1: Rearrange as $y' = 2xy - 12y + 3x - 18 = y(2x - 12) + 3(x - 6)$. Step 2: Write differential equation as: $$\frac{dy}{dx} - (2x - 12)y = 3(x - 6)$$ Step 3: Recognize this as linear ODE. Find integrating factor: $$\mu(x) = e^{-\int (2x - 12) dx} = e^{- (x^{2} - 12x)} = e^{12x - x^{2}}$$ Step 4: Multiply both sides by $\mu(x)$: $$\frac{d}{dx}[y e^{12x - x^{2}}] = 3(x-6) e^{12x - x^{2}}$$ Step 5: Integrate RHS: $$y e^{12x - x^{2}} = \int 3(x-6) e^{12x - x^{2}} dx + C$$ (This integral has no elementary closed form. Leave as integral expression.) 3. Solve $y' = \cos(2x)$, with $y = \sin(2x)$ passing through $(0, \pi/4)$. Step 1: Integrate $y' = \cos(2x)$: $$y = \int \cos(2x) dx = \frac{1}{2} \sin(2x) + C$$ Step 2: Use $y(0) = \pi/4$: $$\pi/4 = \frac{1}{2}\sin(0) + C = 0 + C \Rightarrow C = \frac{\pi}{4}$$ Step 3: Solution: $$y = \frac{1}{2}\sin(2x) + \frac{\pi}{4}$$ 4. Solve $y' = x^{2} e^{x}$. Step 1: Integrate: $$y = \int x^{2} e^{x} dx + C$$ Step 2: Use integration by parts twice: Let $u = x^{2}$, $dv = e^{x} dx$, then $du = 2x dx$, $v = e^{x}$: $$\int x^{2} e^{x} dx = x^{2} e^{x} - \int 2x e^{x} dx$$ Repeat for $\int 2x e^{x} dx$: Let $u = 2x$, $dv = e^{x} dx$, then $du = 2 dx$, $v = e^{x}$: $$\int 2x e^{x} dx = 2x e^{x} - \int 2 e^{x} dx = 2x e^{x} - 2 e^{x} + C$$ Step 3: Substitute back: $$y = x^{2} e^{x} - (2x e^{x} - 2 e^{x}) + C = e^{x}(x^{2} - 2x + 2) + C$$ 5. Solve differential equation $(x^{2} + 9x + 8) dy = dx$. Step 1: Write as: $$dy = \frac{dx}{x^{2} + 9x + 8}$$ Step 2: Integrate both sides: $$y = \int \frac{dx}{x^{2} + 9x + 8} + C$$ Step 3: Factor denominator: $$x^{2} + 9x + 8 = (x + 1)(x + 8)$$ Step 4: Partial fraction decomposition: $$\frac{1}{(x+1)(x+8)} = \frac{A}{x+1} + \frac{B}{x+8}$$ Solving, $A = \frac{1}{7}$, $B = -\frac{1}{7}$. Step 5: Integrate: $$y = \frac{1}{7} \ln|x+1| - \frac{1}{7} \ln|x+8| + C = \frac{1}{7} \ln\left|\frac{x+1}{x+8}\right| + C$$ II. Homogeneous Differential Equations 1. Determine if $(x - 2y) dx - (2x + y) dy = 0$ is homogeneous. Step 1: Rewrite as: $$M = x - 2y, \quad N = -(2x + y)$$ Step 2: Check if $M$ and $N$ are homogeneous functions of same degree. Both are degree 1 (linear in $x,y$), so homogeneous. Step 3: Use substitution $v = y/x$ to solve: Rewrite differential form: $$\frac{dy}{dx} = \frac{M}{N} = \frac{x-2y}{-(2x + y)}$$ Replace $y = vx$, differentiate: $$\frac{dy}{dx} = v + x \frac{dv}{dx}$$ Step 4: Substitute and solve for $dv/dx$ leading to separable differential equation. Step 5: Solve, integrate, and back-substitute for $y$. (not fully expanded here to keep concise) 2. Check $y^{2} dx + (x - y) dy = 0$. Step 1: $M = y^{2}$, $N = x - y$. Degrees: $M$ degree 2, $N$ degree 1. Step 2: Not homogeneous (degrees differ). 3. Check $y' = \frac{x^{2} + y^{2}}{xy}$. Step 1: RHS numerator degree 2, denominator degree 2, so function is homogeneous of degree 0. Step 2: Substitute $v = y/x$, transform equation and solve separable differential. 4. $y' = 2(2x^{2} + y^{2}) dx - xy dy = 0$ (rewrite first). Step 1: Given expression ambiguous but treating as implicit form: $$2(2x^{2} + y^{2}) dx - xy dy = 0$$ Step 2: $M = 2(2x^{2} + y^{2}) = 4x^{2} + 2y^{2}$, $N = -xy$. Degrees: $M$ degree 2, $N$ degree 2. Step 3: Homogeneous. Use substitution $v = y/x$ and solve. 5. $y' = x^{2} + 4 / y$ Step 1: RHS is $x^{2} + \frac{4}{y}$ not homogeneous (sum of terms with different degrees). Note: Solutions to homogeneous DEs involve substitution $v = y/x$ followed by separable integration. Complete solutions might require lengthy integrations or implicit forms.