Question

Solve the differential equation \( ydx - (x + 2y^2) dy = 0 \).

Hint:

If a differential equation is not linear in the form \( \frac{dy}{dx} \), always check if it becomes linear by treating x as the dependent variable and y as the independent variable, i.e., by arranging it into the form \( \frac{dx}{dy} \).

Answer 1

Step 1: Understanding the Concept:
This differential equation is not separable or homogeneous. We can rearrange it to check if it's a linear differential equation. A linear differential equation can be written in the form \( \frac{dy}{dx} + P(x)y = Q(x) \) or \( \frac{dx}{dy} + P(y)x = Q(y) \).
Step 2: Key Formula or Approach:
1. Rearrange the equation into the standard linear form.
2. Identify the functions \( P(y) \) and \( Q(y) \).
3. Calculate the integrating factor (I.F.): \( e^{\int P(y) dy} \).
4. The general solution is given by \( x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.}) dy + C \).
Step 3: Detailed Explanation or Calculation:
The given equation is \( ydx - (x + 2y^2) dy = 0 \).
Rearrange it as:
\[ ydx = (x + 2y^2) dy \] \[ \frac{dx}{dy} = \frac{x + 2y^2}{y} = \frac{x}{y} + 2y \] \[ \frac{dx}{dy} - \frac{1}{y}x = 2y \] This is a linear differential equation in x, with \( P(y) = -\frac{1}{y} \) and \( Q(y) = 2y \).
Now, find the integrating factor:
\[ \text{I.F.} = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = e^{\ln|y|^{-1}} = \frac{1}{|y|} \] Assuming \( y>0 \), the I.F. is \( \frac{1}{y} \).
The general solution is:
\[ x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.}) dy + C \] \[ x \cdot \frac{1}{y} = \int (2y) \left(\frac{1}{y}\right) dy + C \] \[ \frac{x}{y} = \int 2 \, dy + C \] \[ \frac{x}{y} = 2y + C \] \[ x = y(2y + C) = 2y^2 + Cy \] Step 4: Final Answer:
The general solution of the differential equation is \( x = 2y^2 + Cy \).