The straight line `y=2x` meets `y=f(x)` at P, where f(x) is a solution of the differential equation `(dy)/(dx)=(x^(2)+xy)/(x^(2)+y^(2))` such that `f(1)=3`, then `f'(x)` at point P is

Options

A. `(3)/(5)`
B. `(5)/(3)`
C. 2
D. 1

Answer

To solve the problem, we need to find the value of \( f'(x) \) at the point \( P \) where the line \( y = 2x \) intersects the function \( y = f(x) \), given that \( f(x) \) is a solution to the differential equation \( \frac{dy}{dx} = \frac{x^2 + xy}{x^2 + y^2} \) and \( f(1) = 3 \).

### Step-by-step Solution:

1. **Identify the intersection point \( P \)**:
Since the line \( y = 2x \) meets the curve \( y = f(x) \) at point \( P \), we can denote the coordinates of point \( P \) as \( (a, 2a) \).

2. **Substitute into the differential equation**:
We know that at point \( P \), \( y = f(a) = 2a \). We can substitute \( y = 2a \) into the differential equation:
\[ \frac{dy}{dx} = \frac{x^2 + xy}{x^2 + y^2} \]
becomes:
\[ f'(a) = \frac{a^2 + a(2a)}{a^2 + (2a)^2} \]

3. **Simplify the expression**:
Now, simplify the right-hand side:
\[ f'(a) = \frac{a^2 + 2a^2}{a^2 + 4a^2} = \frac{3a^2}{5a^2} \]
Here, \( a^2 \) cancels out (assuming \( a \neq 0 \)):
\[ f'(a) = \frac{3}{5} \]

4. **Conclusion**:
Therefore, the value of \( f'(x) \) at point \( P \) is:
\[ f'(x) = \frac{3}{5} \]

### Final Answer:
The value of \( f'(x) \) at point \( P \) is \( \frac{3}{5} \).

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