Solve the following differential equations.
` (dy)/(dx) = (4x+y+1)^2`

Answer

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

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

### Step 2: Differentiate
Now, differentiate both sides with respect to \( x \):
\[ \frac{dy}{dx} = \frac{dz}{dx} - 4 \]

### Step 3: Substitute into the original equation
Substituting \(\frac{dy}{dx}\) into the original equation gives:
\[ \frac{dz}{dx} - 4 = z^2 \]
Rearranging this, we have:
\[ \frac{dz}{dx} = z^2 + 4 \]

### Step 4: Separation of Variables
Now, we can separate the variables:
\[ \frac{dz}{z^2 + 4} = dx \]

### Step 5: Integrate both sides
Next, we integrate both sides. The left side can be simplified using the formula for the integral of \(\frac{1}{x^2 + a^2}\):
\[ \int \frac{dz}{z^2 + 4} = \int \frac{dz}{z^2 + 2^2} = \frac{1}{2} \tan^{-1}\left(\frac{z}{2}\right) + C \]
The right side integrates to:
\[ \int dx = x + C_1 \]

### Step 6: Combine the results
Setting the two integrals equal gives:
\[ \frac{1}{2} \tan^{-1}\left(\frac{z}{2}\right) = x + C_1 \]

### Step 7: Solve for \( z \)
Multiplying through by 2:
\[ \tan^{-1}\left(\frac{z}{2}\right) = 2x + 2C_1 \]
Taking the tangent of both sides:
\[ \frac{z}{2} = \tan(2x + 2C_1) \]
Thus,
\[ z = 2 \tan(2x + 2C_1) \]

### Step 8: Substitute back for \( y \)
Recall that \( z = 4x + y + 1 \):
\[ 4x + y + 1 = 2 \tan(2x + 2C_1) \]
Rearranging gives:
\[ y = 2 \tan(2x + 2C_1) - 4x - 1 \]

### Final Solution
Thus, the solution to the differential equation is:
\[ y = 2 \tan(2x + C) - 4x - 1 \]
where \( C = 2C_1 \) is an arbitrary constant.