Question

If \( \frac{dy}{dx} = \frac{2}{x+y} \) and \( y(1)=0 \), then \( x+y+2 \) equals

• A. \(3e^{y/2}\)
• B. \(2e^{y/2}\)
• C. \(e^{y/2}\)
• D. \(0\)
• E. \(5e^{y/2}\)

Hint:

Use substitution when variables appear as a sum.

Answer 1

The Correct Option is A

Concept: Use substitution \(v = x+y\) to reduce differential equation.

Step 1: Substitute \(v = x+y\). \[ \frac{dv}{dx} = 1 + \frac{dy}{dx} \] Given: \[ \frac{dy}{dx} = \frac{2}{x+y} = \frac{2}{v} \] Thus: \[ \frac{dv}{dx} = 1 + \frac{2}{v} \]

Step 2: Separate variables. \[ \frac{v}{v+2} dv = dx \]

Step 3: Integrate both sides. \[ \int \frac{v}{v+2} dv = \int dx \] \[ \frac{v}{v+2} = 1 - \frac{2}{v+2} \] \[ \int \left(1 - \frac{2}{v+2}\right) dv = x + C \] \[ v - 2\ln|v+2| = x + C \]

Step 4: Substitute back \(v=x+y\). \[ x+y - 2\ln(x+y+2) = x + C \] \[ y - 2\ln(x+y+2) = C \]

Step 5: Apply initial condition. At \(x=1, y=0\): \[ 0 - 2\ln(3) = C \] \[ C = -2\ln 3 \]

Step 6: Final form. \[ y - 2\ln(x+y+2) = -2\ln 3 \] \[ 2\ln(x+y+2) = y + 2\ln 3 \] \[ \ln(x+y+2)^2 = \ln(3^2 e^y) \] \[ (x+y+2)^2 = 9e^y \] \[ x+y+2 = 3e^{y/2} \]