Question

If the degree of the differential equation corresponding to the family of curves
\[ y = ax + \frac{1}{a} \quad (\text{where } a \neq 0 \text{ is an arbitrary constant}) \] is \(r\) and its order is \(m\), then the solution of
\[ \frac{dy}{dx} - \frac{y}{2x}, \quad y(1) = \sqrt{r + m} \] is

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

Hint:

Identify the degree and order of the differential equation from the given family by eliminating constants. Use initial conditions after solving the DE to find specific solutions.

Answer 1

The Correct Option is B

The given family of curves is: \[ y = ax + \frac{1}{a} \] This can be written as: \[ a^2 y = a^3 x + 1 \Rightarrow a^2 y - a^3 x = 1 \] Differentiating both sides repeatedly will lead to elimination of the constant \(a\). Since we need to eliminate one arbitrary constant, the order is \(m = 1\). Because we ultimately eliminate the constant by differentiating once, and since the highest power of the derivative is 1, the degree is \(r = 2\) (after eliminating square root terms). Now using the differential equation: \[ \frac{dy}{dx} = \frac{y}{2x} \] This is a separable equation: \[ \frac{dy}{y} = \frac{dx}{2x} \Rightarrow \ln y = \frac{1}{2} \ln x + \ln C \Rightarrow \ln y = \ln (C x^{1/2}) \Rightarrow y = C \sqrt{x} \] Now use the initial condition: \[ y(1) = \sqrt{r + m} = \sqrt{2 + 1} = \sqrt{3} \Rightarrow \sqrt{3} = C \Rightarrow C = \sqrt{3} \] Therefore, \[ y = \sqrt{3} \sqrt{x} \Rightarrow y^2 = 3x \]