Find the particular solutions of the following differential equations : ` (dx)/(x+2) + (dy)/(y+2) = 0 , " when " x =1 , y =2 `

Answer

To find the particular solution of the differential equation

\[ \frac{dx}{x+2} + \frac{dy}{y+2} = 0 \]

given the initial conditions \( x = 1 \) and \( y = 2 \), we will follow these steps:

### Step 1: Separate the Variables
We can rearrange the equation to separate the variables:

\[ \frac{dx}{x+2} = -\frac{dy}{y+2} \]

### Step 2: Integrate Both Sides
Next, we integrate both sides:

\[ \int \frac{dx}{x+2} = -\int \frac{dy}{y+2} \]

The integrals yield:

\[ \ln|x+2| = -\ln|y+2| + C \]

where \( C \) is the constant of integration.

### Step 3: Simplify the Equation
We can rewrite the equation using properties of logarithms:

\[ \ln|x+2| + \ln|y+2| = C \]

This can be combined into a single logarithm:

\[ \ln|(x+2)(y+2)| = C \]

### Step 4: Exponentiate to Remove the Logarithm
Exponentiating both sides gives:

\[ (x+2)(y+2) = e^C \]

Let \( k = e^C \), so we have:

\[ (x+2)(y+2) = k \]

### Step 5: Use Initial Conditions to Find \( k \)
Now we will use the initial conditions \( x = 1 \) and \( y = 2 \):

\[ (1+2)(2+2) = k \]

Calculating this gives:

\[ 3 \cdot 4 = k \implies k = 12 \]

### Step 6: Write the Particular Solution
Thus, the particular solution is:

\[ (x+2)(y+2) = 12 \]

### Step 7: Expand and Rearrange
Expanding this gives:

\[ xy + 2x + 2y + 4 = 12 \]

Rearranging leads to:

\[ xy + 2x + 2y - 8 = 0 \]

This is the particular solution of the given differential equation.