Question

Solve the following differential equation ;
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.