Question

Solve the differential equation \( (x + 2y^3) \frac{dy{dx} = y \).}

Hint:

If a differential equation isn't linear in \( y \), try flipping it to see if it's linear in \( x \) (\(dx/dy\) form).
Always remember that \( e^{\log f(y)} = f(y) \), but handle coefficients like the minus sign in \( -\log y \) by moving them inside the log as exponents first.

Answer 1

Concept:
• Linear Differential Equation of the form \( \frac{dx}{dy} + Px = Q \), where \( P \) and \( Q \) are functions of \( y \).
• Integrating Factor (I.F.) = \( e^{\int P \, dy} \).
• General Solution: \( x \cdot (I.F.) = \int Q \cdot (I.F.) \, dy + C \).

Step 1: Rearrange the equation to a standard linear form
The given equation is \( (x + 2y^3) \frac{dy}{dx} = y \). Since it is not linear in \( y \), let's check for linearity in \( x \) by taking the reciprocal: \[ \frac{dx}{dy} = \frac{x + 2y^3}{y} \] \[ \frac{dx}{dy} = \frac{x}{y} + 2y^2 \] \[ \frac{dx}{dy} - \frac{1}{y}x = 2y^2 \] This is a linear differential equation in \( x \) with \( P = -\frac{1}{y} \) and \( Q = 2y^2 \).

Step 2: Calculate the Integrating Factor (I.F.)
\[ I.F. = e^{\int P \, dy} = e^{\int -\frac{1}{y} \, dy} \] \[ I.F. = e^{-\log |y|} = e^{\log |y^{-1}|} \] \[ I.F. = \frac{1}{y} \]

Step 3: Find the general solution
The solution is given by: \[ x \cdot \frac{1}{y} = \int (2y^2) \cdot \frac{1}{y} \, dy \] \[ \frac{x}{y} = \int 2y \, dy \] \[ \frac{x}{y} = y^2 + C \] Multiplying through by \( y \): \[ x = y^3 + Cy \]