Question

Solve: \(ydx - (x + 2y^2)dy = 0\).

Hint:

If a differential equation is not easily separable or linear in the form \(\frac{dy}{dx}\), try rearranging it into the form \(\frac{dx}{dy}\). Sometimes, the equation is linear with respect to the other variable.

Answer 1

Step 1: Understanding the Concept:
This is a first-order differential equation. It is not immediately separable, but we can rearrange it to see if it fits the form of a linear differential equation. By rewriting it with \(x\) as the dependent variable and \(y\) as the independent variable (\(\frac{dx}{dy}\)), we can identify it as a linear equation of the form \(\frac{dx}{dy} + P(y)x = Q(y)\).

Step 2: Key Formula or Approach:
1. Rearrange the equation into the standard linear form \(\frac{dx}{dy} + P(y)x = Q(y)\).
2. Identify \(P(y)\) and \(Q(y)\).
3. Calculate the integrating factor (I.F.) using the formula: I.F. = \(e^{\int P(y)\,dy}\).
4. The general solution is given by: \(x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.})\,dy + C\).
5. Solve the integral and find the final solution.

Step 3: Detailed Explanation:
The given equation is:
\[ y\,dx - (x + 2y^2)\,dy = 0 \]
Rearrange to solve for \(\frac{dx}{dy}\):
\[ y\,dx = (x + 2y^2)\,dy \]
\[ \frac{dx}{dy} = \frac{x + 2y^2}{y} = \frac{x}{y} + 2y \]
Arrange this into the standard linear form:
\[ \frac{dx}{dy} - \frac{1}{y}x = 2y \]
This is a linear differential equation in \(x\).
1. Identify \(P(y)\) and \(Q(y):
\(P(y) = -\frac{1}{y}\) and \(Q(y) = 2y\).
2. Calculate the Integrating Factor (I.F.):
\[ \text{I.F.} = e^{\int P(y)\,dy} = e^{\int -\frac{1}{y}\,dy} = e^{-\ln|y|} = \frac{1}{y} \]
(Assuming \(y > 0\), we can drop the absolute value).
3. Apply the solution formula:
\[ x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.})\,dy + C \]
\[ x \cdot \frac{1}{y} = \int (2y) \cdot \left(\frac{1}{y}\right)\,dy + C \]
\[ \frac{x}{y} = \int 2\,dy + C \]
4. Integrate and find the solution:
\[ \frac{x}{y} = 2y + C \]
\[ x = y(2y+C) = 2y^2 + Cy \]

Step 4: Final Answer:
The general solution to the differential equation is \(x = 2y^2 + Cy\), where \(C\) is an arbitrary constant.