Question

The solution of the differential equation \(\frac{dy}{dx} = \frac{x-y}{x+y}\), when \(x = 0\) and \(y = 0\) represents ....

• A. Circle
• B. Ellipse
• C. Hyperbola
• D. Pair of straight Lines

Hint:

This is a homogeneous equation; put y = vx and integrate.

Answer 1

The Correct Option is D

Step 1: Substitute y = vx:
\(\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}=\dfrac{1-v}{1+v}\). So
\[ x\frac{dv}{dx}=\frac{1-v}{1+v}-v=\frac{1-2v-v^2}{1+v} \]

Step 2: Separate the Variables:
\[ \int\frac{1+v}{1-2v-v^2}\,dv=\int\frac{dx}x \]
The numerator is \(-\tfrac12\) times the derivative of the denominator, so \(-\tfrac12\ln|1-2v-v^2|=\ln|x|+c_1\).

Step 3: General Solution:
\(x^2(1-2v-v^2)=C\), and with \(v=y/x\): \(x^2-2xy-y^2=C\).

Step 4: Use the Condition:
The curve passes through \((0,0)\), so \(C=0\) and the solution is \(x^2-2xy-y^2=0\). This factorises into two straight lines through the origin (discriminant \(h^2-ab=1+1=2>0\)).
So the solution is a pair of straight lines, option (D). A circle, ellipse or hyperbola would have a constant term \(C\neq0\).

Final Answer:
The solution represents a pair of straight lines, option (D). \[ \boxed{\text{(D) Pair of straight lines}} \]