Question

Find the general solution of the differential equation: \[ y \, dx = (x + 2y^2) \, dy. \]

Hint:

For equations involving logarithms, utilize properties such as \( \ln a - \ln b = \ln \left(\frac{a}{b}\right) \) to simplify and solve effectively.

Answer 1

We are given a differential equation. Let us solve it step by step. Step 1: Rewrite the equation. The given differential equation is: \[ \frac{dx}{dy} - \frac{x}{y} = 2y \] This is a first-order linear differential equation of the form: \[ \frac{dx}{dy} + P(y)x = Q(y), \] where \(P(y) = -\frac{1}{y}\) and \(Q(y) = 2y\). Step 2: Find the integrating factor (IF). The integrating factor for a linear differential equation is given by: \[ \text{Integrating Factor} = e^{\int P(y) \, dy}. \] Here, \(P(y) = -\frac{1}{y}\), so: \[ \text{Integrating Factor} = e^{\int -\frac{1}{y} \, dy} = e^{-\ln|y|} = \frac{1}{y}. \] Step 3: Solve the equation. Multiply the entire differential equation by the integrating factor \(\frac{1}{y}\): \[ \frac{1}{y} \frac{dx}{dy} - \frac{x}{y^2} = \frac{2y}{y}. \] This simplifies to: \[ \frac{d}{dy}\left(\frac{x}{y}\right) = 2. \] Now integrate both sides with respect to \(y\): \[ \frac{x}{y} = \int 2 \, dy. \] Step 4: Integrate. The integral of \(2 \, dy\) is: \[ \frac{x}{y} = 2y + C, \] where \(C\) is the constant of integration. Step 5: Solve for \(x\). Multiply through by \(y\) to isolate \(x\): \[ x = 2y^2 + Cy. \] Final Answer: The solution to the given differential equation is: \[ \boxed{x = 2y^2 + Cy.} \]