The solution of the differential equation `(dy)/(dx)=(x+y)/(x)` satisfying the condition y (1) = 1 is :

Options

A. `y=log x +x`
B. `y=x log x +x^(2)`
C. `y=x e^((x-1))`
D. `y=x log x +x`

Answer

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

### Step 1: Rewrite the differential equation
We start with the given equation:
\[ \frac{dy}{dx} = \frac{x + y}{x} \]
This can be rewritten as:
\[ \frac{dy}{dx} = 1 + \frac{y}{x} \]

### Step 2: Substitute \(y = vx\)
Let \(y = vx\), where \(v\) is a function of \(x\). Then, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \]

### Step 3: Substitute into the differential equation
Substituting \(y = vx\) into the rewritten differential equation gives:
\[ v + x\frac{dv}{dx} = 1 + v \]
Simplifying this, we find:
\[ x\frac{dv}{dx} = 1 \]

### Step 4: Separate variables
Now we can separate the variables:
\[ dv = \frac{1}{x}dx \]

### Step 5: Integrate both sides
Integrating both sides:
\[ \int dv = \int \frac{1}{x}dx \]
This results in:
\[ v = \log|x| + C \]

### Step 6: Substitute back for \(y\)
Recalling that \(y = vx\), we substitute back:
\[ y = x(\log|x| + C) \]

### Step 7: Apply the initial condition
Now, we use the initial condition \(y(1) = 1\):
\[ 1 = 1(\log|1| + C) \]
Since \(\log|1| = 0\), we have:
\[ 1 = C \]

### Step 8: Write the final solution
Substituting \(C\) back into the equation for \(y\):
\[ y = x(\log|x| + 1) \]
Thus, the final solution is:
\[ y = x \log x + x \]