Question

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 ______.

Answer 1

Correct Answer: 20

The line \( Y - y = Y'(x)(X - x) \) represents the tangent to \( Y = Y(X) \) at \( (x, y) \).

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

Differentiate with respect to \( x \) and solve for \( Y(x) \) using the initial condition \( Y(1) = 1 \).

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

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

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

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

Answer 2

This problem describes a curve \(Y=Y(X)\) in the first quadrant through a property of its tangent line. We must first establish the differential equation governing the curve based on the given area condition, solve it using the initial condition \(Y(1)=1\), and finally calculate the value of \(12Y(2)\).

Concept Used:

1. Equation of a Tangent Line: The equation of the tangent line to the curve \(Y = Y(X)\) at a point \((x, y)\) is given by \(Y_{tan} - y = Y'(x)(X_{tan} - x)\).

2. Intercepts of a Line: The Y-intercept is found by setting \(X_{tan}=0\), and the X-intercept is found by setting \(Y_{tan}=0\).

3. Area of a Triangle: The area of the triangle formed by a line and the coordinate axes is \(A = \frac{1}{2} \times |\text{X-intercept}| \times |\text{Y-intercept}|\).

4. First-Order Linear Differential Equation: A differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \) can be solved using an integrating factor (I.F.) given by \( e^{\int P(x) dx} \). The solution is \( y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C \).

Step-by-Step Solution:

Step 1: Find the intercepts of the tangent line with the coordinate axes.

The equation of the tangent at an arbitrary point \((x, y)\) on the curve is \(Y_{tan} - y = Y'(x)(X_{tan} - x)\).

To find the Y-intercept, set \(X_{tan} = 0\):

\[ Y_{int} - y = Y'(x)(0 - x) \implies Y_{int} = y - xY'(x) \]

To find the X-intercept, set \(Y_{tan} = 0\):

\[ 0 - y = Y'(x)(X_{int} - x) \implies X_{int} = x - \frac{y}{Y'(x)} = \frac{xY'(x) - y}{Y'(x)} \]

Step 2: Calculate the area enclosed by the tangent and the axes.

The area \(A\) of the triangle is:

\[ A = \frac{1}{2} |X_{int} \cdot Y_{int}| = \frac{1}{2} \left| \left( \frac{xY'(x) - y}{Y'(x)} \right) \left( y - xY'(x) \right) \right| \] \[ A = \frac{1}{2} \left| \frac{-(y - xY'(x))^2}{Y'(x)} \right| = \frac{(y - xY'(x))^2}{2|Y'(x)|} \]

Step 3: Formulate the differential equation using the given area.

We are given that the area is \( \frac{-y^2}{2Y'(x)} + 1 \). Since the curve lies in the first quadrant (\(y>0\)) and the area must be positive, the term \( \frac{-y^2}{2Y'(x)} \) implies that \( Y'(x) < 0 \). With \( Y'(x) < 0 \), we have \( |Y'(x)| = -Y'(x) \).

Equating the two expressions for the area:

\[ \frac{(y - xY'(x))^2}{-2Y'(x)} = \frac{-y^2}{2Y'(x)} + 1 \]

Multiply both sides by \( -2Y'(x) \):

\[ (y - xY'(x))^2 = y^2 - 2Y'(x) \] \[ y^2 - 2xyY'(x) + x^2(Y'(x))^2 = y^2 - 2Y'(x) \]

Simplifying the equation:

\[ -2xyY'(x) + x^2(Y'(x))^2 = -2Y'(x) \]

Since \( Y'(x) \neq 0 \), we can divide the entire equation by \( Y'(x) \):

\[ -2xy + x^2Y'(x) = -2 \] \[ x^2Y'(x) - 2xy = -2 \]

Step 4: Solve the first-order linear differential equation.

Rearranging the equation into the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \):

\[ Y' - \frac{2}{x}y = -\frac{2}{x^2} \]

Here, \( P(x) = -\frac{2}{x} \) and \( Q(x) = -\frac{2}{x^2} \). The integrating factor (I.F.) is:

\[ \text{I.F.} = e^{\int P(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2\ln x} = e^{\ln(x^{-2})} = x^{-2} = \frac{1}{x^2} \]

The solution is given by \( y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C \):

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

The general solution for the curve is:

\[ Y(x) = \frac{2}{3x} + Cx^2 \]

Step 5: Use the initial condition \( Y(1) = 1 \) to find the constant \( C \).

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

The particular solution is:

\[ Y(x) = \frac{2}{3x} + \frac{1}{3}x^2 = \frac{2 + x^3}{3x} \]

Final Computation & Result:

We need to find the value of \( 12Y(2) \).

First, calculate \( Y(2) \):

\[ Y(2) = \frac{2 + (2)^3}{3(2)} = \frac{2 + 8}{6} = \frac{10}{6} = \frac{5}{3} \]

Now, compute the final value:

\[ 12Y(2) = 12 \times \frac{5}{3} = 4 \times 5 = 20 \]

The value of \( 12Y(2) \) is 20.