Let's solve the differential equation step by step: Given: \[ \frac{dy}{dx} = \frac{y}{x} + \frac{y^2}{x^3} \] ### Step 1: Rewrite the Equation Factor \( y \) from the right side: \[ \frac{dy}{dx} = \frac{y}{x} + \frac{y^2}{x^3} = \frac{y}{x} + \frac{y^2}{x^3} \] ### Step 2: Get All \( y \)-Terms on One Side This is a first-order nonlinear ODE. Let's write the equation as: \[ \frac{dy}{dx} = \frac{y}{x} + \frac{y^2}{x^3} \] It can be written as: \[ \frac{dy}{dx} - \frac{y}{x} = \frac{y^2}{x^3} \] ### Step 3: Use an Appropriate Substitution Since the right side involves \( y^2 \), this is a **Bernoulli equation** of the form: \[ \frac{dy}{dx} + P(x)y = Q(x) y^n \] Here, \( P(x) = -\frac{1}{x} \), \( Q(x) = \frac{1}{x^3} \), and \( n = 2 \). For Bernoulli equations, use the substitution \( v = y^{1-n} = y^{-1} \). So, let \( v = \frac{1}{y} \). Now, differentiate both sides with respect to \( x \): \[ \frac{dv}{dx} = -\frac{1}{y^2}\frac{dy}{dx} \] So, \[ \frac{dy}{dx} = -y^2 \frac{dv}{dx} \] ### Step 4: Substitute Back into the Original Equation Substitute \( y \) and \( \frac{dy}{dx} \) in terms of \( v \): \[ -y^2 \frac{dv}{dx} - \frac{y}{x} = \frac{y^2}{x^3} \] Divide both sides by \( y^2 \): \[ -\frac{dv}{dx} - \frac{1}{x}\frac{1}{y} = \frac{1}{x^3} \] But \(\frac{1}{y} = v\): \[ -\frac{dv}{dx} - \frac{v}{x} = \frac{1}{x^3} \] Multiply both sides by \(-1\): \[ \frac{dv}{dx} + \frac{v}{x} = -\frac{1}{x^3} \] ### Step 5: This is Now a Linear ODE in \( v \) The equation: \[ \frac{dv}{dx} + \frac{v}{x} = -\frac{1}{x^3} \] This is a first-order linear ODE. #### Find the integrating factor (IF): \[ IF = e^{\int \frac{1}{x} dx} = e^{\ln |x|} = x \] #### Multiply through by the integrating factor: \[ x \frac{dv}{dx} + v = -\frac{1}{x^2} \] Left side is derivative of \( v \cdot x \): \[ \frac{d}{dx}(v x) = -\frac{1}{x^2} \] Integrate both sides: \[ \int \frac{d}{dx}(v x) dx = \int -\frac{1}{x^2} dx \] \[ v x = \frac{1}{x} + C \] (Recall \(\int -x^{-2} dx = x^{-1}\)) ### Step 6: Solve for \( v \): \[ v x = \frac{1}{x} + C \] \[ v = \frac{1}{x^2} + \frac{C}{x} \] Recall \( v = \frac{1}{y} \): \[ \frac{1}{y} = \frac{1}{x^2} + \frac{C}{x} \] ### Step 7: Final Answer \[ \boxed{ \frac{1}{y} = \frac{1}{x^2} + \frac{C}{x} } \] or, equivalently, \[ \boxed{ y = \frac{1}{\frac{1}{x^2} + \frac{C}{x}} } \] --- **Summary of Steps:** 1. Recognize as Bernoulli and substitute \( v = 1/y \). 2. Convert to a linear ODE in \( v \). 3. Solve by integrating factor. 4. Substitute back for \( y \). Let me know if you want more explanation for any step!