`(x+2y +1) dx - (2x + 4y +3)dy = 0, x+2y = u`

Answer

To solve the differential equation \((x + 2y + 1)dx - (2x + 4y + 3)dy = 0\) with the substitution \(x + 2y = u\), we can follow these steps:

### Step 1: Rewrite the differential equation
We start with the given equation:
\[ (x + 2y + 1)dx - (2x + 4y + 3)dy = 0 \]
Rearranging gives us:
\[ (x + 2y + 1)dx = (2x + 4y + 3)dy \]
Now, we can express \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{x + 2y + 1}{2x + 4y + 3} \]

### Step 2: Substitute \(u = x + 2y\)
Using the substitution \(u = x + 2y\), we can express \(y\) in terms of \(u\) and \(x\):
\[ y = \frac{u - x}{2} \]
Now, we need to find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1}{2} \frac{du}{dx} - \frac{1}{2} \]

### Step 3: Substitute \(\frac{dy}{dx}\) into the equation
Substituting \(\frac{dy}{dx}\) back into the equation gives:
\[ \frac{1}{2} \frac{du}{dx} - \frac{1}{2} = \frac{u + 1}{2u + 3} \]
Multiplying through by 2 to eliminate the fraction:
\[ \frac{du}{dx} - 1 = \frac{2u + 2}{2u + 3} \]
Rearranging gives:
\[ \frac{du}{dx} = \frac{2u + 2}{2u + 3} + 1 \]

### Step 4: Simplify the right-hand side
To combine the fractions on the right-hand side, we find a common denominator:
\[ \frac{du}{dx} = \frac{2u + 2 + (2u + 3)}{2u + 3} = \frac{4u + 5}{2u + 3} \]

### Step 5: Separate variables
Now we separate the variables:
\[ \frac{2u + 3}{4u + 5} du = dx \]

### Step 6: Integrate both sides
Integrating both sides:
\[ \int \frac{2u + 3}{4u + 5} du = \int dx \]
To integrate the left side, we can use substitution or partial fractions. Here, we can split the integral:
\[ \int \left( \frac{1}{2} + \frac{1}{4u + 5} \right) du = x + C \]
This gives:
\[ \frac{1}{2}u + \frac{1}{4} \ln |4u + 5| = x + C \]

### Step 7: Substitute back for \(u\)
Substituting back \(u = x + 2y\):
\[ \frac{1}{2}(x + 2y) + \frac{1}{4} \ln |4(x + 2y) + 5| = x + C \]

### Step 8: Simplify the equation
Rearranging gives us the final solution in terms of \(y\):
\[ y + \frac{1}{8} \ln |4x + 8y + 5| = \frac{x}{2} + C \]

### Final Answer
Thus, the solution to the differential equation is:
\[ y + \frac{1}{8} \ln |4x + 8y + 5| = \frac{x}{2} + C \]