Key Concepts and Formulas

• Equation of a Tangent Line: The equation of the tangent line to the curve $Y=Y(X)$ at the point $(x,y)$ is given by $Y - y = Y'(x)(X - x)$.
• Area of a Triangle: The area of a triangle formed by a line and the coordinate axes can be calculated using the x and y intercepts of the line. If the x-intercept is $a$ and the y-intercept is $b$, the area of the triangle is $\frac{1}{2}|ab|$.
• Differential Equations: A differential equation is an equation that relates a function to its derivatives. Solving a differential equation involves finding the function that satisfies the equation.

Step-by-Step Solution

Step 1: Find the x and y intercepts of the tangent line

The equation of the tangent line is given by $Y - y = Y'(x)(X - x)$. We need to find where this line intersects the x and y axes.

• x-intercept (Y=0): $0 - y = Y'(x)(X - x)$ $-y = Y'(x)X - Y'(x)x$ $Y'(x)X = Y'(x)x - y$ $X = x - \frac{y}{Y'(x)}$ So, the x-intercept is $x - \frac{y}{Y'(x)}$.

• y-intercept (X=0): $Y - y = Y'(x)(0 - x)$ $Y = y - xY'(x)$ So, the y-intercept is $y - xY'(x)$.

Step 2: Calculate the area of the triangle formed by the tangent line and the coordinate axes

The area of the triangle formed by the tangent line and the coordinate axes is given by: $Area = \frac{1}{2} \left| \left(x - \frac{y}{Y'(x)}\right) \left(y - xY'(x)\right) \right|$ Since the curve is in the first quadrant and we're told $Y'(x) \neq 0$, we can assume the intercepts are such that the absolute value signs can be dropped, or that we are considering the correct signs for the problem setup. We are given that the area is $Area = \frac{-y^2}{2Y'(x)} + 1$ Therefore: $\frac{1}{2} \left(x - \frac{y}{Y'(x)}\right) \left(y - xY'(x)\right) = \frac{-y^2}{2Y'(x)} + 1$

Step 3: Simplify the equation and form a differential equation

Multiplying both sides by 2: $\left(x - \frac{y}{Y'(x)}\right) \left(y - xY'(x)\right) = \frac{-y^2}{Y'(x)} + 2$ Expanding the left side: $xy - x^2Y'(x) - \frac{y^2}{Y'(x)} + xy = \frac{-y^2}{Y'(x)} + 2$ $2xy - x^2Y'(x) - \frac{y^2}{Y'(x)} = \frac{-y^2}{Y'(x)} + 2$ $2xy - x^2Y'(x) = 2$ $x^2Y'(x) = 2xy - 2$ $Y'(x) = \frac{2xy - 2}{x^2}$ $\frac{dY}{dx} = \frac{2xy - 2}{x^2}$ $\frac{dY}{dx} = \frac{2y}{x} - \frac{2}{x^2}$

Step 4: Solve the differential equation

This is a first-order linear differential equation of the form $\frac{dY}{dx} + P(x)Y = Q(x)$, where $P(x) = -\frac{2}{x}$ and $Q(x) = -\frac{2}{x^2}$.

The integrating factor (IF) is given by: $IF = e^{\int P(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2\ln|x|} = e^{\ln(x^{-2})} = x^{-2} = \frac{1}{x^2}$

The solution is given by: $Y(x) \cdot IF = \int Q(x) \cdot IF \, dx$ $Y(x) \cdot \frac{1}{x^2} = \int -\frac{2}{x^2} \cdot \frac{1}{x^2} \, dx$ $\frac{Y(x)}{x^2} = \int -\frac{2}{x^4} \, dx$ $\frac{Y(x)}{x^2} = \frac{2}{3x^3} + C$ $Y(x) = \frac{2}{3x} + Cx^2$

Step 5: Apply the initial condition Y(1) = 1

We are given that $Y(1) = 1$. Substituting this into the equation: $1 = \frac{2}{3(1)} + C(1)^2$ $1 = \frac{2}{3} + C$ $C = 1 - \frac{2}{3} = \frac{1}{3}$

So, the solution is: $Y(x) = \frac{2}{3x} + \frac{1}{3}x^2$

Step 6: Find Y(2) and calculate 12Y(2)

We need to find $Y(2)$: $Y(2) = \frac{2}{3(2)} + \frac{1}{3}(2)^2 = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}$

Now, we calculate $12Y(2)$: $12Y(2) = 12 \cdot \frac{5}{3} = 4 \cdot 5 = 20$

Common Mistakes & Tips

• Sign Errors: Be careful with the signs when calculating the intercepts and the area.
• Integrating Factor: Remember to correctly calculate the integrating factor and use it properly in the solution.
• Algebraic Simplification: Carefully simplify the equations to avoid errors.

Summary

We found the equation of the tangent line, calculated the area of the triangle formed by the tangent and the coordinate axes, and set it equal to the given expression. This led to a first-order linear differential equation, which we solved using the integrating factor method. Applying the initial condition $Y(1) = 1$ allowed us to find the specific solution $Y(x) = \frac{2}{3x} + \frac{1}{3}x^2$. Finally, we calculated $Y(2)$ and found that $12Y(2) = 20$.

The final answer is \boxed{20}.