Solve the following differential equation ;

Question

y'= 2y'x + 1/y

Answer 1

To solve the differential equation \( y' = 2y' x + \frac{1}{y} \), let's start by rearranging it to isolate \( y' \).

1. Rearranging the Equation: We can express this as: \[ y' - 2y' x = \frac{1}{y} \] Factoring out \( y' \) gives: \[ y'(1 - 2x) = \frac{1}{y} \] Therefore, we have: \[ y' = \frac{1}{y(1 - 2x)} \]

2. Using Separation of Variables: We can separate the variables: \[ y , dy = \frac{1}{1 - 2x} , dx \]

3. Integrating Both Sides: Now we integrate both sides: \[ \int y , dy = \int \frac{1}{1 - 2x} , dx \] The left side integrates to: \[ \frac{y^2}{2} + C_1 \] The right side integrates to: \[ -\frac{1}{2} \ln |1 - 2x| + C_2 \]

4. Combining the Results: Setting the constant terms together: \[ \frac{y^2}{2} = -\frac{1}{2} \ln |1 - 2x| + C \]

5. Multiplying Through by 2 (to simplify): \[ y^2 = -\ln |1 - 2x| + 2C \]

Alternatively, we can let \( C = 2C_1 \): \[ y^2 = -\ln |1 - 2x| + C \]

This equation describes the solution to the original differential equation in terms of \( y^{2} \) in relation to \( x \).

6. Final Rearrangement: To express \( y \) explicitly, we can write: \[ y = \sqrt{-\ln |1 - 2x| + C} \]

This represents the general solution of the given differential equation, where \( C \) is an arbitrary constant.