Question

Find the solution of \((x^2-3y^2)dx+2xydy=0\)

• A. \(x^2-y^2=kx^2\)
• B. \(y^2-x^2=kx^3\)
• C. \(y^2-x^2=-kx^3\)
• D. \(x^2-y^2=-kx^2\)

Hint:

For homogeneous differential equations, use \(y=vx\), then substitute \(\frac{dy}{dx}=v+x\frac{dv}{dx}\).

Answer 1

The Correct Option is B

Concept:
The given differential equation is: \[ (x^2-3y^2)dx+2xydy=0 \] This is a homogeneous differential equation because every term has the same degree in \(x\) and \(y\). So, we use the substitution: \[ y=vx \] Then: \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \]

Step 1: Convert the equation into derivative form.
\[ (x^2-3y^2)+2xy\frac{dy}{dx}=0 \] \[ 2xy\frac{dy}{dx}=3y^2-x^2 \]

Step 2: Put \(y=vx\).
\[ y=vx \] So: \[ y^2=v^2x^2 \] and: \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Substitute in: \[ (x^2-3y^2)+2xy\frac{dy}{dx}=0 \] \[ x^2-3v^2x^2+2x(vx)\left(v+x\frac{dv}{dx}\right)=0 \] \[ x^2(1-3v^2)+2vx^2\left(v+x\frac{dv}{dx}\right)=0 \] \[ 1-3v^2+2v^2+2vx\frac{dv}{dx}=0 \] \[ 1-v^2+2vx\frac{dv}{dx}=0 \] \[ 2vx\frac{dv}{dx}=v^2-1 \]

Step 3: Separate the variables.
\[ \frac{2v}{v^2-1}dv=\frac{dx}{x} \]

Step 4: Integrate both sides.
\[ \int \frac{2v}{v^2-1}dv=\int \frac{dx}{x} \] \[ \log|v^2-1|=\log|x|+\log k \] \[ v^2-1=kx \]

Step 5: Substitute \(v=\frac{y}{x}\).
\[ \left(\frac{y}{x}\right)^2-1=kx \] \[ \frac{y^2}{x^2}-1=kx \] \[ \frac{y^2-x^2}{x^2}=kx \] \[ y^2-x^2=kx^3 \] Hence, the correct answer is: \[ \boxed{(B)\ y^2-x^2=kx^3} \]