1. Problem a) Solve the Bernoulli differential equation $$\frac{dy}{dx} - x y = e^{-x^2} y^3$$ Step 1: Identify the Bernoulli form: $$\frac{dy}{dx} + P(x) y = Q(x) y^n$$ with $$P(x) = -x$$, $$Q(x) = e^{-x^2}$$, and $$n=3$$. Step 2: Substitute $$v = y^{1-n} = y^{-2}$$, then $$\frac{dv}{dx} = -2 y^{-3} \frac{dy}{dx}$$. Step 3: Rewrite the original equation: $$\frac{dy}{dx} = x y + e^{-x^2} y^3$$ Multiply both sides by $$-2 y^{-3}$$: $$\frac{dv}{dx} = -2 x y^{-2} - 2 e^{-x^2}$$ Since $$v = y^{-2}$$, this becomes: $$\frac{dv}{dx} + 2 x v = -2 e^{-x^2}$$ Step 4: Solve the linear ODE for $$v$$ using integrating factor: $$\mu(x) = e^{\int 2x dx} = e^{x^2}$$ Multiply both sides by $$\mu(x)$$: $$e^{x^2} \frac{dv}{dx} + 2 x e^{x^2} v = -2 e^{x^2} e^{-x^2} = -2$$ Left side is derivative: $$\frac{d}{dx} (v e^{x^2}) = -2$$ Step 5: Integrate both sides: $$v e^{x^2} = -2x + C$$ Step 6: Solve for $$v$$: $$v = e^{-x^2} (-2x + C)$$ Step 7: Recall $$v = y^{-2}$$, so: $$y = \pm \frac{1}{\sqrt{e^{-x^2} (-2x + C)}} = \pm \frac{e^{x^2/2}}{\sqrt{C - 2x}}$$ --- 2. Problem b) Solve $$\frac{dy}{dx} = \frac{y}{x} + x y^2$$ Step 1: Rewrite as: $$\frac{dy}{dx} - \frac{y}{x} = x y^2$$ Step 2: Identify Bernoulli form with $$P(x) = -\frac{1}{x}$$, $$Q(x) = x$$, $$n=2$$. Step 3: Substitute $$v = y^{1-2} = y^{-1}$$, then $$\frac{dv}{dx} = - y^{-2} \frac{dy}{dx}$$. Step 4: Multiply original equation by $$-y^{-2}$$: $$- y^{-2} \frac{dy}{dx} + \frac{y^{-1}}{x} = - x$$ Rewrite: $$\frac{dv}{dx} + \frac{1}{x} v = - x$$ Step 5: Solve linear ODE for $$v$$ with integrating factor: $$\mu(x) = e^{\int \frac{1}{x} dx} = x$$ Multiply both sides: $$x \frac{dv}{dx} + v = - x^2$$ Left side is derivative: $$\frac{d}{dx} (v x) = - x^2$$ Step 6: Integrate: $$v x = - \frac{x^3}{3} + C$$ Step 7: Solve for $$v$$: $$v = - \frac{x^2}{3} + \frac{C}{x}$$ Step 8: Recall $$v = y^{-1}$$, so: $$y = \frac{1}{v} = \frac{1}{- \frac{x^2}{3} + \frac{C}{x}} = \frac{x}{C - \frac{x^3}{3}} = \frac{3x}{3C - x^3}$$ --- 3. Problem c) Solve $$\frac{dy}{dx} - \frac{3}{2x} y = \frac{2x}{y}$$ Step 1: Rewrite as Bernoulli form: $$\frac{dy}{dx} + P(x) y = Q(x) y^n$$ with $$P(x) = -\frac{3}{2x}$$, $$Q(x) = 2x$$, $$n = -1$$ (since $$\frac{2x}{y} = 2x y^{-1}$$). Step 2: Substitute $$v = y^{1 - (-1)} = y^2$$, then $$\frac{dv}{dx} = 2 y \frac{dy}{dx}$$. Step 3: Multiply original equation by $$2 y$$: $$2 y \frac{dy}{dx} - \frac{3}{x} y^2 = 4 x$$ Rewrite: $$\frac{dv}{dx} - \frac{3}{x} v = 4 x$$ Step 4: Solve linear ODE for $$v$$ with integrating factor: $$\mu(x) = e^{\int -\frac{3}{x} dx} = e^{-3 \ln x} = x^{-3}$$ Multiply both sides: $$x^{-3} \frac{dv}{dx} - \frac{3}{x} x^{-3} v = 4 x x^{-3} = 4 x^{-2}$$ Left side is derivative: $$\frac{d}{dx} (v x^{-3}) = 4 x^{-2}$$ Step 5: Integrate: $$v x^{-3} = -4 x^{-1} + C$$ Step 6: Solve for $$v$$: $$v = x^{3} (-4 x^{-1} + C) = -4 x^{2} + C x^{3}$$ Step 7: Recall $$v = y^2$$, so: $$y = \pm \sqrt{-4 x^{2} + C x^{3}} = \pm x \sqrt{C x - 4}$$ --- 4. Problem d) Solve $$x y^{2} \frac{dy}{dx} + y^{3} = 1$$ Step 1: Rewrite: $$x y^{2} \frac{dy}{dx} = 1 - y^{3}$$ Step 2: Divide both sides by $$x y^{2}$$: $$\frac{dy}{dx} = \frac{1}{x y^{2}} - \frac{y}{x}$$ Step 3: Substitute $$v = y^{-1}$$, then $$\frac{dv}{dx} = - y^{-2} \frac{dy}{dx}$$. Step 4: Multiply both sides by $$- y^{-2}$$: $$- y^{-2} \frac{dy}{dx} = - \frac{1}{x y^{4}} + \frac{y}{x y^{2}} = - \frac{1}{x y^{4}} + \frac{1}{x y}$$ Rewrite in terms of $$v$$: $$\frac{dv}{dx} = - \frac{1}{x} v^{4} + \frac{1}{x} v^{-1}$$ Step 5: This is nonlinear and complicated; instead, multiply original equation by $$y^{-3}$$: $$x y^{-1} \frac{dy}{dx} + 1 = y^{-3}$$ Step 6: Substitute $$v = y^{-1}$$, so $$\frac{dv}{dx} = - y^{-2} \frac{dy}{dx}$$, then: $$\frac{dy}{dx} = - y^{2} \frac{dv}{dx} = - \frac{1}{v^{2}} \frac{dv}{dx}$$ Step 7: Substitute into equation: $$x y^{-1} \frac{dy}{dx} + 1 = y^{-3}$$ $$x v (- \frac{1}{v^{2}} \frac{dv}{dx}) + 1 = v^{3}$$ Simplify: $$- \frac{x}{v} \frac{dv}{dx} + 1 = v^{3}$$ Rearranged: $$- \frac{x}{v} \frac{dv}{dx} = v^{3} - 1$$ Multiply both sides by $$- \frac{v}{x}$$: $$\frac{dv}{dx} = \frac{v}{x} (1 - v^{3})$$ Step 8: Separate variables: $$\frac{dv}{v (1 - v^{3})} = \frac{dx}{x}$$ Step 9: Partial fraction decomposition for left side: $$\frac{1}{v (1 - v^{3})} = \frac{1}{v (1 - v)(1 + v + v^{2})}$$ Step 10: Integrate both sides: $$\int \frac{dv}{v (1 - v^{3})} = \int \frac{dx}{x} = \ln |x| + C$$ Step 11: The integral on left is complex; final implicit solution is: $$\int \frac{dv}{v (1 - v^{3})} = \ln |x| + C$$ Step 12: Recall $$v = y^{-1}$$, so solution is implicit in terms of $$y$$ and $$x$$. --- Final answers: a) $$y = \pm \frac{e^{x^{2}/2}}{\sqrt{C - 2x}}$$ b) $$y = \frac{3x}{3C - x^{3}}$$ c) $$y = \pm x \sqrt{C x - 4}$$ d) Implicit solution: $$\int \frac{dv}{v (1 - v^{3})} = \ln |x| + C$$ with $$v = y^{-1}$$