Question

Which of the following is the solution of \[ p^2+p\left(\frac{y}{x}-\frac{x}{y}\right)-1=0, \] where \[ p=\frac{dy}{dx}\,? \]

• A. \((xy-c)(x^2+y^2-c)=0\)
• B. \(\left(\dfrac{x}{y}-c\right)(x^2-y^2+c)=0\)
• C. \((xy-c)(x^2-y^2-c)=0\)
• D. \((xy-cx)(x^2+y^2-c)=0\)

Hint:

When a differential equation is quadratic in \(p=\frac{dy}{dx}\), factorize it and solve each first-order equation separately.

Answer 1

The Correct Option is C

Concept:
Here, \[ p=\frac{dy}{dx} \] The differential equation is quadratic in \(p\): \[ p^2+p\left(\frac{y}{x}-\frac{x}{y}\right)-1=0 \] We solve it by factorizing as a quadratic equation in \(p\).

Step 1: Write the equation. \[ p^2+p\left(\frac{y}{x}-\frac{x}{y}\right)-1=0 \] The middle coefficient is \[ \frac{y}{x}-\frac{x}{y} \]

Step 2: Factorize the quadratic. Observe that \[ p^2+p\left(\frac{y}{x}-\frac{x}{y}\right)-1 \] can be written as \[ \left(p+\frac{y}{x}\right)\left(p-\frac{x}{y}\right)=0 \] because \[ \left(p+\frac{y}{x}\right)\left(p-\frac{x}{y}\right) = p^2+p\left(\frac{y}{x}-\frac{x}{y}\right)-1 \]

Step 3: Solve first factor. \[ p+\frac{y}{x}=0 \] \[ \frac{dy}{dx}=-\frac{y}{x} \] Separate variables: \[ \frac{dy}{y}=-\frac{dx}{x} \] Integrate: \[ \log y=-\log x+\log c \] \[ \log(xy)=\log c \] \[ xy=c \] So one factor of the complete solution is \[ xy-c=0 \]

Step 4: Solve second factor. \[ p-\frac{x}{y}=0 \] \[ \frac{dy}{dx}=\frac{x}{y} \] So, \[ y\,dy=x\,dx \] Integrate: \[ \frac{y^2}{2}=\frac{x^2}{2}+C \] \[ y^2=x^2+C_1 \] \[ x^2-y^2=C \] So the second factor is \[ x^2-y^2-c=0 \]

Step 5: Combine both solutions. The complete solution is written as \[ (xy-c)(x^2-y^2-c)=0 \]

Step 6: Final answer. \[ \boxed{(xy-c)(x^2-y^2-c)=0} \]