The solution of the differential equation `(dy)/(dx)=(4x+y+1)^(2)`, is

Options

A. `4x +y+1=2 tan (2x +y+C)`
B. `4x + y +1 =2 tan (x+2y +C)`
C. `4x +y +1 =2 tan (2y +c)`
D. `4x +y +1 =2 tan (2x +c)`

Answer

To solve the differential equation \(\frac{dy}{dx} = (4x + y + 1)^2\), we will follow these steps:

### Step 1: Substitute for simplicity
Let \(t = 4x + y + 1\). Then, we can express \(y\) in terms of \(t\) and \(x\):
\[ y = t - 4x - 1 \]

### Step 2: Differentiate \(t\) with respect to \(x\)
Now, we differentiate \(t\) with respect to \(x\):
\[ \frac{dt}{dx} = 4 + \frac{dy}{dx} \]

### Step 3: Substitute \(\frac{dy}{dx}\) into the equation
From the original differential equation, we know:
\[ \frac{dy}{dx} = t^2 \]
Substituting this into our expression for \(\frac{dt}{dx}\):
\[ \frac{dt}{dx} = 4 + t^2 \]

### Step 4: Separate variables
We can now separate the variables:
\[ \frac{dt}{4 + t^2} = dx \]

### Step 5: Integrate both sides
Now we integrate both sides:
\[ \int \frac{dt}{4 + t^2} = \int dx \]
The left side can be integrated using the formula \(\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C\). Here, \(a = 2\):
\[ \frac{1}{2} \tan^{-1} \left(\frac{t}{2}\right) = x + C \]

### Step 6: Solve for \(t\)
Multiplying both sides by 2 gives:
\[ \tan^{-1} \left(\frac{t}{2}\right) = 2x + 2C \]
Taking the tangent of both sides:
\[ \frac{t}{2} = \tan(2x + 2C) \]
Thus:
\[ t = 2 \tan(2x + 2C) \]

### Step 7: Substitute back for \(t\)
Now, substituting back for \(t\):
\[ 4x + y + 1 = 2 \tan(2x + 2C) \]
Rearranging gives:
\[ y = 2 \tan(2x + 2C) - 4x - 1 \]

### Final Answer
Thus, the solution of the differential equation is:
\[ y = 2 \tan(2x + C') - 4x - 1 \]
where \(C' = 2C\) is a constant.