Question

Let $y=y(x)$ be the solution of the differential equation $\left(x^2-3 y^2\right) d x+3 x y d y=0, y(1)=1$.Then $6 y^2( e )$ is equal to

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

Answer 1

The Correct Option is D

The correct answer is (D) : $2 e ^2$
$\left(x^2-3y^2\right)dx+3xydy=0$
$\frac{dy}{dx}=\frac{3y^2-x^2}{3xy}\Rightarrow \frac{dy}{dx}=\frac{y}{x}-\frac{1}{3}\frac{x}{y}.....(1)$
Put $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
(1) $\Rightarrow v+x\frac{dv}{dx}=v-\frac{1}{3}\frac{1}{v}$
$\Rightarrow vdv=\frac{-1}{3x}$
Integrating both side
$\frac{v^2}{2}=\frac{-1}{3}\ln x+c$
$\Rightarrow \frac{y^2}{2x^2}=\frac{-1}{3}\ln x+c$
$y(1)=1$
$\Rightarrow \frac{1}{2}=c$
$\Rightarrow \frac{y^2}{2x^2}=\frac{-1}{3}\ln x+\frac{1}{2}$
$\Rightarrow y^2=-\frac{2}{3}{x}^2\ln x+x^2$
$y^2(e)=-\frac{2}{3}{e}^2+e^2=\frac{e^2}{3}$
$\Rightarrow 6y^2(e)=2e^2$

Answer 2

Step 1: Use the substitution \( y = vx \) 

The given differential equation is:

\[ (x^2 - 3y^2)dx + 3xy \, dy = 0. \] Rearrange the equation: \[ \frac{dy}{dx} = \frac{3y^2 - x^2}{3xy}. \]

Now, use the substitution \( y = vx \), so \( \frac{dy}{dx} = v + x \frac{dv}{dx} \). Substituting into the equation:

\[ v + x \frac{dv}{dx} = \frac{3(vx)^2 - x^2}{3vx} = \frac{x^2(3v^2 - 1)}{3vx}. \] Simplify: \[ v + x \frac{dv}{dx} = \frac{3v^2 - 1}{3v}. \] 

Step 2: Solve for \( \frac{dv}{dx} \) \[ x \frac{dv}{dx} = \frac{3v^2 - 1}{3v} - v = \frac{3v^2 - 1 - 3v^2}{3v} = -\frac{1}{3v}. \] Thus: \[ \frac{dv}{dx} = -\frac{1}{3vx}. \] 

Step 3: Solve the differential equation Separate the variables: \[ v \, dv = -\frac{1}{3x} \, dx. \] Integrate both sides: \[ \int v \, dv = -\frac{1}{3} \int \frac{1}{x} \, dx. \] The integrals give: \[ \frac{v^2}{2} = -\frac{1}{3} \ln x + C. \] 

Step 4: Substitute \( v = \frac{y}{x} \) \[ \frac{y^2}{2x^2} = -\frac{1}{3} \ln x + C. \] Multiply through by 2: \[ y^2 = 2x^2 \left(-\frac{1}{3} \ln x + C \right). \] 

Step 5: Apply the initial condition \( y(1) = 1 \) Substitute \( x = 1 \) and \( y = 1 \) into the equation: \[ 1^2 = 2(1)^2 \left(-\frac{1}{3} \ln 1 + C \right). \] Since \( \ln 1 = 0 \), this simplifies to: \[ 1 = 2C \quad \Rightarrow \quad C = \frac{1}{2}. \] 

Step 6: Find \( 6y^2(e) \) Substitute \( C = \frac{1}{2} \) and \( x = e \) into the equation: \[ y^2 = 2e^2 \left(-\frac{1}{3} \ln e + \frac{1}{2} \right). \] Since \( \ln e = 1 \), we get: \[ y^2 = 2e^2 \left(-\frac{1}{3} + \frac{1}{2} \right). \] Simplify: \[ y^2 = 2e^2 \left(-\frac{2}{6} + \frac{3}{6} \right) = 2e^2 \cdot \frac{1}{6} = \frac{e^2}{3}. \] Multiply by 6: \[ 6y^2 = 6 \cdot \frac{e^2}{3} = 2e^2. \]