Solve the following differential equations :
`(x+y+1)^(2)dy=dx, y(-1)=0`.

Answer

To solve the differential equation \((x+y+1)^{2} dy = dx\) with the initial condition \(y(-1) = 0\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ (x+y+1)^{2} dy = dx \]
We can rearrange it to express \(\frac{dx}{dy}\):
\[ \frac{dx}{dy} = (x+y+1)^{2} \]

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

### Step 3: Substitute into the equation
Substituting \(\frac{dx}{dy}\) into our rearranged equation gives:
\[ \frac{dv}{dy} - 1 = v^{2} \]
Rearranging this, we have:
\[ \frac{dv}{dy} = v^{2} + 1 \]

### Step 4: Separate variables
We can separate variables to integrate:
\[ \frac{dv}{v^{2} + 1} = dy \]

### Step 5: Integrate both sides
Now, we integrate both sides:
\[ \int \frac{dv}{v^{2} + 1} = \int dy \]
The integral of \(\frac{1}{v^{2} + 1}\) is \(\tan^{-1}(v)\), and the integral of \(dy\) is \(y + C\):
\[ \tan^{-1}(v) = y + C \]

### Step 6: Substitute back for \(v\)
Substituting back for \(v\):
\[ \tan^{-1}(x + y + 1) = y + C \]

### Step 7: Apply the initial condition
Using the initial condition \(y(-1) = 0\):
\[ \tan^{-1}(-1 + 0 + 1) = 0 + C \]
This simplifies to:
\[ \tan^{-1}(0) = C \implies C = 0 \]

### Step 8: Final solution
Thus, we have:
\[ \tan^{-1}(x + y + 1) = y \]
This can be rewritten as:
\[ x + y + 1 = \tan(y) \]
So the final solution is:
\[ x + y + 1 = \tan(y) \]