Solution of differential equation `e^((dy)/(dx))=x` is

Options

A. `xlog|x|-1-y=c`
B. `xlog|x|-1+y=c`
C. `x(log|x|-1)+y=c`
D. `x(log|x|-1)-y=c`

Answer

To solve the differential equation \( e^{\frac{dy}{dx}} = x \), we will follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides of the equation:
\[ \frac{dy}{dx} = \ln(x) \]

### Step 2: Separate the variables
Next, we separate the variables \( y \) and \( x \):
\[ dy = \ln(x) \, dx \]

### Step 3: Integrate both sides
Now, we will integrate both sides. The left side integrates to \( y \), and the right side requires integration by parts:
\[ \int dy = \int \ln(x) \, dx \]
For the right side, we will use integration by parts where we let:
- \( u = \ln(x) \) (thus \( du = \frac{1}{x} \, dx \))
- \( dv = dx \) (thus \( v = x \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int \ln(x) \, dx = x \ln(x) - \int x \cdot \frac{1}{x} \, dx = x \ln(x) - \int 1 \, dx = x \ln(x) - x + C \]

### Step 4: Write the solution
Putting it all together, we have:
\[ y = x \ln(x) - x + C \]

### Final Answer
Thus, the solution to the differential equation \( e^{\frac{dy}{dx}} = x \) is:
\[ y = x \ln(x) - x + C \]

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