Solve : `(dy)/(dx)= (2x+3y-4)^(2)`.

Answer

To solve the differential equation \(\frac{dy}{dx} = (2x + 3y - 4)^2\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \frac{dy}{dx} = (2x + 3y - 4)^2 \]

### Step 2: Introduce a substitution
Let \(v = 2x + 3y - 4\). Then, we can express \(y\) in terms of \(v\):
\[ 3y = v + 4 - 2x \implies y = \frac{v + 4 - 2x}{3} \]
Now, differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{1}{3} \frac{dv}{dx} - \frac{2}{3} \]

### Step 3: Substitute back into the original equation
Substituting \(\frac{dy}{dx}\) into the original equation gives:
\[ \frac{1}{3} \frac{dv}{dx} - \frac{2}{3} = v^2 \]
Multiplying through by 3 to eliminate the fraction:
\[ \frac{dv}{dx} - 2 = 3v^2 \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ \frac{dv}{dx} = 3v^2 + 2 \]

### Step 5: Separate variables
Now we can separate the variables:
\[ \frac{dv}{3v^2 + 2} = dx \]

### Step 6: Integrate both sides
Integrate both sides:
\[ \int \frac{dv}{3v^2 + 2} = \int dx \]
The left side can be integrated using a standard integral formula. The integral of \(\frac{1}{a^2 + x^2}\) is \(\frac{1}{a} \tan^{-1}(\frac{x}{a})\).

Let \(a^2 = \frac{2}{3}\) or \(a = \sqrt{\frac{2}{3}}\):
\[ \int \frac{dv}{3v^2 + 2} = \frac{1}{\sqrt{6}} \tan^{-1}\left(\frac{v\sqrt{3}}{\sqrt{2}}\right) + C \]
The right side integrates to:
\[ x + C_1 \]

### Step 7: Combine results
Thus, we have:
\[ \frac{1}{\sqrt{6}} \tan^{-1}\left(\frac{v\sqrt{3}}{\sqrt{2}}\right) = x + C \]

### Step 8: Substitute back for \(v\)
Substituting back \(v = 2x + 3y - 4\):
\[ \frac{1}{\sqrt{6}} \tan^{-1}\left(\frac{(2x + 3y - 4)\sqrt{3}}{\sqrt{2}}\right) = x + C \]

### Final Result
This gives us the implicit solution to the differential equation:
\[ \frac{1}{\sqrt{6}} \tan^{-1}\left(\frac{(2x + 3y - 4)\sqrt{3}}{\sqrt{2}}\right) = x + C \]