Question

Let \( x = x(y) \) be the solution of the differential equation} \[ 2y^2 \frac{dx}{dy} - 2xy + x^2 = 0, \quad y>1, \quad x(e) = e. \] Then \( x(e^2) \) is equal to:

• A. \( \frac{3}{2} e^2 \)
• B. \( \frac{2}{3} e^2 \)
• C. \( e^2 \)
• D. \( 2e^2 \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are given a first-order non-linear differential equation.
By rearranging the terms and dividing by \(y^2\), we can check if it is a homogeneous differential equation.
Once verified, the standard substitution \(x = vy\) will transform it into a separable differential equation.
Step 2: Key Formula or Approach:
Rewrite the equation by dividing entirely by \(y^2\).
\[ 2 \frac{dx}{dy} - 2\frac{x}{y} + \left(\frac{x}{y}\right)^2 = 0 \] This is clearly a homogeneous equation of degree zero.
Let \(x = vy\), which implies \(\frac{dx}{dy} = v + y\frac{dv}{dy}\).
Step 3: Detailed Explanation:
Substitute \(x/y = v\) and \(\frac{dx}{dy}\) into the modified equation.
\[ 2\left(v + y\frac{dv}{dy}\right) - 2v + v^2 = 0 \] Distribute the 2.
\[ 2v + 2y\frac{dv}{dy} - 2v + v^2 = 0 \] The \(2v\) terms cancel out perfectly.
\[ 2y\frac{dv}{dy} + v^2 = 0 \implies 2y\frac{dv}{dy} = -v^2 \] Now separate the variables \(v\) and \(y\).
\[ \frac{-2}{v^2} dv = \frac{dy}{y} \] Integrate both sides.
\[ \int -2v^{-2} dv = \int \frac{1}{y} dy \] \[ \frac{2}{v} = \ln|y| + C \] Substitute back \(v = \frac{x}{y}\).
\[ \frac{2y}{x} = \ln y + C \quad (\text{since } y>1) \] Use the initial condition \(x(e) = e\) to find the constant \(C\).
\[ \frac{2e}{e} = \ln(e) + C \implies 2 = 1 + C \implies C = 1 \] So, the particular solution curve is:
\[ \frac{2y}{x} = \ln y + 1 \] Rearrange to express \(x\) as a function of \(y\).
\[ x = \frac{2y}{\ln y + 1} \] Finally, evaluate \(x\) at \(y = e^2\).
\[ x(e^2) = \frac{2e^2}{\ln(e^2) + 1} \] \[ x(e^2) = \frac{2e^2}{2\ln(e) + 1} = \frac{2e^2}{2(1) + 1} = \frac{2e^2}{3} \] Step 4: Final Answer:
The value of \(x(e^2)\) is \(\frac{2}{3} e^2\).