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: Form the equation
\(\frac{dy}{dx} = \frac{y}{x+y}\), so \(\frac{dx}{dy} = \frac{x+y}{y} = \frac xy + 1\).

Step 2: Make it linear
\(\frac{dx}{dy} - \frac1y x = 1\). The integrating factor is \(e^{-\int dy/y} = \frac1y\).

Step 3: Solve
\(\frac{d}{dy}\left(\frac xy\right) = \frac1y\), so \(\frac xy = \ln y + c\).

Step 4: Use the point
At \((0,1)\): \(0 = 0 + c\), so \(c=0\). Then \(\frac xy = \ln y\), that is \(y = e^{x/y}\). Option (D).

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