Limit Evaluation Steps

The given limit is:

$ \lim_{t \to x} \left( \frac{t^2 y(x) - x^2 y(t)}{x - t} \right) = 3 $

This limit takes the indeterminate form $0/0$ as $t$ approaches $x$. We apply L'Hôpital's Rule, differentiating the numerator and denominator with respect to $t$:

• Numerator derivative: $\frac{d}{dt} (t^2 y(x) - x^2 y(t)) = 2t y(x) - x^2 y'(t)$
• Denominator derivative: $\frac{d}{dt} (x - t) = -1$

Applying the rule and evaluating the limit at $t = x$:

$ \frac{2x y(x) - x^2 y'(x)}{-1} = x^2 y'(x) - 2x y(x) $

Equating this expression to 3 yields the differential equation:

$ x^2 y'(x) - 2x y(x) = 3 $

Differential Equation Solution

Rewrite the equation in standard linear form:

$ y'(x) - \frac{2}{x} y(x) = \frac{3}{x^2} $

This is a first-order linear ordinary differential equation (ODE). Calculate the integrating factor $I(x)$:

$ I(x) = e^{\int -\frac{2}{x} dx} = e^{-2 \ln|x|} $

Since the domain is $(0, \infty)$, $|x| = x$:

$ I(x) = e^{-2 \ln x} = e^{\ln(x^{-2})} = x^{-2} = \frac{1}{x^2} $

Multiply the ODE by the integrating factor $I(x)$:

$ \frac{1}{x^2} y'(x) - \frac{2}{x^3} y(x) = \frac{3}{x^4} $

The left side is the derivative of the product $I(x)y(x)$:

$ \frac{d}{dx} \left( \frac{1}{x^2} y(x) \right) = \frac{3}{x^4} $

Integrate both sides with respect to $x$:

$ \frac{1}{x^2} y(x) = \int \frac{3}{x^4} dx = 3 \int x^{-4} dx $ $ \frac{1}{x^2} y(x) = 3 \left( \frac{x^{-3}}{-3} \right) + C = -x^{-3} + C $

Solve for the general solution $y(x)$:

$ y(x) = x^2 \left( -\frac{1}{x^3} + C \right) = -\frac{1}{x} + C x^2 $

Initial Condition Application

Use the given initial condition $y(1) = 2$ to find the constant $C$:

$ 2 = -\frac{1}{1} + C (1)^2 $ $ 2 = -1 + C \implies C = 3 $

Substitute $C = 3$ back into the general solution to get the specific solution:

$ y(x) = -\frac{1}{x} + 3x^2 $

Final Value Calculation

Evaluate the function at $x = 2$:

$ y(2) = -\frac{1}{2} + 3(2)^2 = -\frac{1}{2} + 3(4) = -\frac{1}{2} + 12 $ $ y(2) = \frac{-1 + 24}{2} = \frac{23}{2} $

Calculate the required value $2y(2)$:

$ 2y(2) = 2 \times \frac{23}{2} = 23 $