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 will follow these steps:

### Step 1: Substitute and Rearrange
Let us define a new variable:
\[ u = 4x + y + 1 \]
Then, differentiating \( u \) with respect to \( x \):
\[ \frac{du}{dx} = 4 + \frac{dy}{dx} \]
From the original equation, we can express \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = u^2 \]
Substituting this into the equation for \( \frac{du}{dx} \):
\[ \frac{du}{dx} = 4 + u^2 \]

### Step 2: Separate Variables
Now, we can rearrange the equation:
\[ \frac{du}{u^2 + 4} = dx \]

### Step 3: Integrate Both Sides
Next, we integrate both sides. The left side requires the integral of \( \frac{1}{u^2 + 4} \):
\[ \int \frac{du}{u^2 + 4} = \frac{1}{2} \tan^{-1} \left( \frac{u}{2} \right) + C \]
The right side integrates to:
\[ \int dx = x + C_1 \]
Combining these results, we have:
\[ \frac{1}{2} \tan^{-1} \left( \frac{u}{2} \right) = x + C \]

### Step 4: Solve for \( u \)
To isolate \( u \), we multiply both sides by 2:
\[ \tan^{-1} \left( \frac{u}{2} \right) = 2x + C \]
Taking the tangent of both sides gives:
\[ \frac{u}{2} = \tan(2x + C) \]
Thus, we can express \( u \) as:
\[ u = 2 \tan(2x + C) \]

### Step 5: Substitute Back for \( y \)
Recalling our substitution \( u = 4x + y + 1 \), we substitute back:
\[ 4x + y + 1 = 2 \tan(2x + C) \]
Rearranging gives us the final solution:
\[ y = 2 \tan(2x + C) - 4x - 1 \]

### Final Solution
The solution to the differential equation is:
\[ y = 2 \tan(2x + C) - 4x - 1 \]