Let's solve the given Bernoulli differential equation in **2 steps** with explanations: ### Given: \[ y' + 4xy = xy^3 \] --- ## **Step 1: Convert to Linear Form** First, rewrite the equation: \[ y' + 4xy = xy^3 \] Divide both sides by \( y^3 \): \[ y' y^{-3} + 4x y y^{-3} = x \] \[ y' y^{-3} + 4x y^{-2} = x \] Let \( v = y^{-2} \), so \( \frac{dv}{dx} = -2y^{-3}y' \) or \( y' y^{-3} = -\frac{1}{2} \frac{dv}{dx} \). Substitute into the equation: \[ -\frac{1}{2} \frac{dv}{dx} + 4x v = x \] Multiply both sides by \(-2\): \[ \frac{dv}{dx} - 8x v = -2x \] **Explanation:** We used the substitution \( v = y^{-2} \) to convert the Bernoulli equation to a linear first-order ODE in terms of \( v \). --- ## **Step 2: Solve the Linear ODE** The linear ODE is: \[ \frac{dv}{dx} - 8x v = -2x \] This is of the form \( \frac{dv}{dx} + P(x)v = Q(x) \) where \( P(x) = -8x \), \( Q(x) = -2x \). The integrating factor (IF) is: \[ IF = e^{\int -8x dx} = e^{-4x^2} \] Multiply through by the integrating factor: \[ e^{-4x^2} \frac{dv}{dx} - 8x e^{-4x^2} v = -2x e^{-4x^2} \] \[ \frac{d}{dx} (v e^{-4x^2}) = -2x e^{-4x^2} \] Integrate both sides: \[ v e^{-4x^2} = \int -2x e^{-4x^2} dx + C \] Let \( u = -4x^2 \Rightarrow du = -8x dx \Rightarrow -\frac{1}{8} du = x dx \): So, \[ \int -2x e^{-4x^2} dx = -2 \int x e^{-4x^2} dx = -2 \left(-\frac{1}{8} \int e^{u} du \right) = \frac{1}{4} e^{-4x^2} \] So, \[ v e^{-4x^2} = \frac{1}{4} e^{-4x^2} + C \] \[ v = \frac{1}{4} + Ce^{4x^2} \] Recall \( v = y^{-2} \), so: \[ y^{-2} = \frac{1}{4} + Ce^{4x^2} \] \[ \boxed{y(x) = \left( \frac{1}{4} + Ce^{4x^2} \right)^{-\frac{1}{2}}} \] **Explanation:** We solved the linear ODE using an integrating factor and substituted back to get the solution in terms of \( y(x) \). --- ## **Final Answer:** \[ \boxed{y(x) = \left( \frac{1}{4} + Ce^{4x^2} \right)^{-\frac{1}{2}}} \]