Question

The equation of the curve passing through the point (0, 1) and having a slope of the tangent at point \(P(x,y)\) is equal to \(\frac{y}{x+y}\), is

• A. \(x = e^{-\frac{y}{x}}\)
• B. \(y = e^{-\frac{y}{x}}+1\)
• C. \(x = e^{\frac{x}{y}}-1\)
• D. \(y = e^{\frac{x}{y}}\)

Hint:

Treat x as a function of y to get a linear equation.

Answer 1

The Correct Option is D

Step 1: Substitution
Put $x = vy$. Then $\frac{dx}{dy} = v + y\frac{dv}{dy}$, and the equation $\frac{dx}{dy} = v+1$ gives $y\frac{dv}{dy} = 1$.

Step 2: Integrate
$v = \ln y + c$, so $\frac xy = \ln y + c$. The point $(0,1)$ gives $c=0$.

Step 3: Result
$y = e^{x/y}$, option (D).

Final Answer:
y = e^(x/y). \[ \boxed{\text{(D)}\ y=e^{\frac xy}} \]