If `(dy)/(dx)-x tan (y-x)=1`, then its solution is ……..

Answer

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

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

### Step 2: Substitute into the original equation
Substituting \(y\) and \(\frac{dy}{dx}\) into the differential equation gives:
\[ \frac{dv}{dx} + 1 - x \tan(v) = 1 \]
This simplifies to:
\[ \frac{dv}{dx} - x \tan(v) = 0 \]

### Step 3: Rearrange the equation
Rearranging the equation, we have:
\[ \frac{dv}{dx} = x \tan(v) \]

### Step 4: Separate variables
Now we can separate the variables:
\[ \frac{1}{\tan(v)} dv = x \, dx \]
This can be rewritten as:
\[ \cot(v) \, dv = x \, dx \]

### Step 5: Integrate both sides
Integrating both sides:
\[ \int \cot(v) \, dv = \int x \, dx \]
The left side integrates to:
\[ \log(\sin(v)) + C_1 \]
The right side integrates to:
\[ \frac{x^2}{2} + C_2 \]
Thus, we have:
\[ \log(\sin(v)) = \frac{x^2}{2} + C \]
where \(C = C_2 - C_1\).

### Step 6: Solve for \(\sin(v)\)
Exponentiating both sides:
\[ \sin(v) = e^{\frac{x^2}{2} + C} = e^C e^{\frac{x^2}{2}} \]
Let \(k = e^C\), so:
\[ \sin(v) = k e^{\frac{x^2}{2}} \]

### Step 7: Substitute back for \(v\)
Recall that \(v = y - x\):
\[ \sin(y - x) = k e^{\frac{x^2}{2}} \]

### Final Answer
Thus, the solution to the differential equation is:
\[ \sin(y - x) = k e^{\frac{x^2}{2}} \]