The substitution $y = v x$ transforms the homogeneous differential equation:
$\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$
into the separable differential equation $\frac{2v}{1 - v^2} \, dv = \frac{dx}{x}$. Given the initial condition $y(1) = 0$, the value of $y^2$ when $x = 2$ is ________ (answer in integer).
Answer Key:
2
Step-by-Step Solution & Explanation
Substitute $y = vx \implies \frac{dy}{dx} = v + x \frac{dv}{dx}$:
$v + x \frac{dv}{dx} = \frac{x^2 + v^2 x^2}{2x(vx)} = \frac{1 + v^2}{2v}$
$x \frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 - v^2}{2v}$
Separating variables:
$\frac{2v}{1 - v^2} \, dv = \frac{dx}{x}$
Integrating both sides:
$-\ln|1 - v^2| = \ln|x| + C_1 \implies \ln|(1 - v^2)x| = C \implies (1 - v^2)x = C$
Back-substituting $v = y/x$:
$\left( 1 - \frac{y^2}{x^2} \right) x = C \implies \frac{x^2 - y^2}{x} = C \implies x^2 - y^2 = C x$
Applying the initial condition $y(1) = 0$:
$1^2 - 0^2 = C(1) \implies C = 1$
Thus the particular solution is $x^2 - y^2 = x \implies y^2 = x^2 - x$.
At $x = 2$:
$y^2 = 2^2 - 2 = 4 - 2 = 2$