Solution of differential equation `(dt)/(dx)=(xlogx)/(t)` is

Options

A. `x=e^(-ct)`
B. `x=e^(ct)`
C. `log|x|=e^(-ct)`
D. `log|x|=e^(ct)`

Answer

To solve the differential equation \(\frac{dt}{dx} = \frac{x \log x}{t}\), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to separate the variables \(t\) and \(x\):
\[ t \, dt = x \log x \, dx \]

### Step 2: Integrating Both Sides
Next, we integrate both sides:
\[ \int t \, dt = \int x \log x \, dx \]

### Step 3: Solving the Left Side
The left side integrates easily:
\[ \int t \, dt = \frac{t^2}{2} + C_1 \]

### Step 4: Solving the Right Side
For the right side, we will use integration by parts. Let:
- \(u = \log x\) (then \(du = \frac{1}{x} dx\))
- \(dv = x \, dx\) (then \(v = \frac{x^2}{2}\))

Using integration by parts:
\[ \int x \log x \, dx = u v - \int v \, du \]
Substituting the values:
\[ = \log x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \]
\[ = \frac{x^2 \log x}{2} - \frac{1}{2} \int x \, dx \]
\[ = \frac{x^2 \log x}{2} - \frac{1}{2} \cdot \frac{x^2}{2} + C_2 \]
\[ = \frac{x^2 \log x}{2} - \frac{x^2}{4} + C_2 \]

### Step 5: Equating Both Integrals
Now we equate both sides:
\[ \frac{t^2}{2} = \frac{x^2 \log x}{2} - \frac{x^2}{4} + C \]
where \(C = C_2 - C_1\).

### Step 6: Simplifying the Equation
To simplify, we multiply through by 2:
\[ t^2 = x^2 \log x - \frac{x^2}{2} + 2C \]

### Step 7: Final Form
Thus, the final solution of the differential equation is:
\[ t^2 = x^2 \log x - \frac{x^2}{2} + C' \]
where \(C' = 2C\).