Let $ Y = Y(X) $ be a curve lying in the first quadrant such that the area enclosed by the line $ Y - y = Y'(x) (X - x) $ and the coordinate axes, where $ (x, y) $ is any point on the curve, is always\[\frac{-y^2}{2Y'(x)} + 1, \quad Y'(x) \neq 0.\]If $ Y(1) = 1 $, then $ 12Y(2) $ equals ______.

Question

Given that the solution of \[ \frac{d^2y}{dx^2} + \alpha \frac{dy}{dx} + \beta y = -e^{-x} \] is \[ y(x) = C_1 e^{-x} + C_2 e^{2x} + x e^{-x}, \] find the values of $\alpha$ and $\beta$.

Answer 1

The equation $ Y - y = Y'(x)(X - x) $ defines the tangent to the curve $ Y = Y(X) $ at the point $ (x, y) $.

The relationship $ A = -\frac{Y(x)^2}{2Y'(x)} + 1 $ connects $ y $ and $ Y'(x) $ on the curve.

Differentiate this relationship with respect to $ x $ and then solve for $ Y(x) $, using the initial condition $ Y(1) = 1 $.

\[ 1 = \frac{2}{3} + c \]

\[ c = \frac{1}{3} \]

\[ Y = \frac{2}{3} \times \frac{1}{X} + \frac{1}{3}X^2 \]

\[ 12Y(2) = \frac{5}{3} \times 12 = 20 \]

Correct Answer: 20

Solution and Explanation

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