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Step-by-Step Solution
Step 1: Identify the Tangent Line at a General Point
Given the curve Y = Y(X) in the first quadrant, at any point (x, y) on this curve, the tangent line is:
Y - y = Y'(x)\,(X - x).
Step 2: Find the Intercepts of the Tangent Line
x-intercept: Set Y = 0 in the tangent line
0 - y = Y'(x)\,(X - x)\;\Longrightarrow\;X - x = -\frac{y}{Y'(x)}\;\Longrightarrow\;X = x - \frac{y}{Y'(x)}.
The x-intercept is thus \Bigl(x - \frac{y}{Y'(x)},\,0\Bigr).
y-intercept: Set X = 0 in the tangent line
Y - y = Y'(x)\,(0 - x)\;\Longrightarrow\;Y = y - x\,Y'(x).
The y-intercept is thus \bigl(0,\,y - x\,Y'(x)\bigr).
Step 3: Area of the Triangle Formed by These Intercepts
The triangle formed by these intercepts and the coordinate axes has area:
A = \frac{1}{2} \times \bigl[\text{x-intercept}\bigr] \times \bigl[\text{y-intercept}\bigr].
Hence,
A \;=\; \frac{1}{2} \left(x - \frac{y}{Y'(x)}\right)\bigl(y - x\,Y'(x)\bigr).
Step 4: Simplify the Area and Apply the Given Condition
Expand and simplify:
\begin{aligned}
A &= \frac{1}{2}\Bigl[x\,y - x^2\,Y'(x) \;-\; \frac{y}{Y'(x)}\,y + \frac{y}{Y'(x)}\,x\,Y'(x)\Bigr] \\
&= \frac{1}{2}\Bigl[xy - x^2\,Y'(x) - \frac{y^2}{Y'(x)} + xy\Bigr] \\
&= \frac{1}{2}\Bigl[-x^2\,Y'(x) + 2xy - \frac{y^2}{Y'(x)}\Bigr].
\end{aligned}
We are given that this area A is always \displaystyle \frac{-\,y^2}{2\,Y'(x)} + 1. Hence,
-\frac{1}{2}x^2\,Y'(x) + x\,y - \frac{y^2}{2\,Y'(x)} \;=\; \frac{-y^2}{2\,Y'(x)} + 1.
Observe the terms -\frac{y^2}{2\,Y'(x)} cancel on both sides, leaving:
-\frac{1}{2}x^2\,Y'(x) + x\,y \;=\; 1.
Rearrange to isolate Y'(x) :
x\,y - 1 \;=\; \frac{1}{2}x^2\,Y'(x)
\;\Longrightarrow\; Y'(x) \;=\; \frac{2}{x^2}\,(x\,y - 1).
Alternatively,
Y'(x) = \frac{2\,x\,y - 2}{x^2} = \frac{2\,y}{x} - \frac{2}{x^2}.
Step 5: Form the Differential Equation
We now have:
\frac{dy}{dx} = \frac{2\,y}{x} - \frac{2}{x^2}.
Rewrite it as:
\frac{dy}{dx} - \frac{2}{x}\,y = -\frac{2}{x^2}.
This is a linear first-order differential equation in standard form:
\frac{dy}{dx} + P(x)\,y = Q(x),
with P(x) = -\frac{2}{x} and Q(x) = -\frac{2}{x^2}.
Step 6: Find the Integrating Factor
The integrating factor (IF) is given by:
\displaystyle \text{IF} = \exp\!\Bigl(\int P(x)\,dx\Bigr)
= \exp\!\Bigl(\int -\tfrac{2}{x}\,dx\Bigr)
= \exp(-2 \ln x)
= x^{-2}.
Step 7: Multiply the DE by the Integrating Factor
Multiply the entire differential equation
\frac{dy}{dx} - \frac{2}{x}\,y = -\frac{2}{x^2}
by x^{-2} :
x^{-2}\,\frac{dy}{dx} \;-\;\frac{2}{x}\,x^{-2}\,y \;=\; -\frac{2}{x^2}\,x^{-2}.
The left side simplifies to \frac{d}{dx}[y\,x^{-2}] , and the right side is -\frac{2}{x^4}. Thus,
\frac{d}{dx}\bigl(y\,x^{-2}\bigr) = -\frac{2}{x^4}.
Step 8: Integrate Both Sides
Integrate with respect to x :
y\,x^{-2}
= \int -\frac{2}{x^4}\,dx
= -2\int x^{-4}\,dx
= -2\left(-\frac{1}{3}x^{-3}\right)
= \frac{2}{3}\,x^{-3} + C.
Hence,
y = x^{2}\left(\frac{2}{3}x^{-3} + C\right) = \frac{2}{3}\,\frac{1}{x} + C\,x^{2}.
Step 9: Apply the Initial Condition Y(1) = 1
Given Y(1) = 1, substitute x = 1 :
1 = \frac{2}{3}\,(1)^{-1} + C\,(1)^2 = \frac{2}{3} + C.
Thus, C = 1 - \frac{2}{3} = \frac{1}{3}.
The particular solution becomes:
Y(x) = \frac{2}{3\,x} + \frac{1}{3}\,x^2.
Step 10: Evaluate 12\,Y(2)
Substitute x = 2 :
Y(2) = \frac{2}{3 \times 2} + \frac{1}{3}\,(2)^2
= \frac{1}{3} + \frac{4}{3}
= \frac{5}{3}.
Therefore:
12\,Y(2) = 12 \times \frac{5}{3} = 20.
Final Answer: 20 .
