Question

For all $ x>0 $, let $ y_1(x), y_2(x), y_3(x) $ be the functions satisfying $$ \frac{dy_1}{dx} - (\sin x)^2 y_1 = 0, \quad y_1(1) = 5, $$ $$ \frac{dy_2}{dx} - (\cos x)^2 y_2 = 0, \quad y_2(1) = \frac{1}{3}, $$ $$ \frac{dy_3}{dx} - \left(\frac{2 - x^3}{x^3}\right) y_3 = 0, \quad y_3(1) = \frac{3}{5e}, $$ respectively. Then $$ \lim_{x \to 0^+} \frac{y_1(x)y_2(x)y_3(x) + 2x}{e^{3x} \sin x} $$ is equal to __________.

Hint:

In limits involving ODE solutions and exponential forms, reduce each function separately, simplify using Taylor approximations near 0, and evaluate dominant terms.

Answer 1

Correct Answer: 2

Step 1: Solve the differential equations

We have three differential equations for \( x>0 \), with initial conditions at \( x = 1 \):

1. \(\frac{dy_1}{dx} - (\sin x)^2 y_1 = 0\), \( y_1(1) = 5 \),

2. \(\frac{dy_2}{dx} - (\cos x)^2 y_2 = 0\), \( y_2(1) = \frac{1}{3} \),

3. \(\frac{dy_3}{dx} - \left( \frac{2 - x^3}{x^3} \right) y_3 = 0\), \( y_3(1) = \frac{3}{5e} \).

These are all first-order linear differential equations of the form \(\frac{dy}{dx} + P(x) y = 0\), with solution \( y = Ce^{-\int P(x) \, dx} \). Let’s solve each one.
Solve for \( y_1(x) \): \[ \frac{dy_1}{dx} - (\sin x)^2 y_1 = 0 \implies \frac{dy_1}{dx} = (\sin x)^2 y_1 \implies \frac{dy_1}{y_1} = (\sin x)^2 \, dx. \] Integrate both sides: \[ \ln |y_1| = \int (\sin x)^2 \, dx. \] Use the identity \( (\sin x)^2 = \frac{1 - \cos 2x}{2} \): \[ \int (\sin x)^2 \, dx = \int \frac{1 - \cos 2x}{2} \, dx = \frac{1}{2} \left( x - \frac{\sin 2x}{2} \right) + C_1 = \frac{x}{2} - \frac{\sin 2x}{4} + C_1. \] So: \[ \ln y_1 = \frac{x}{2} - \frac{\sin 2x}{4} + C_1 \quad (\text{since } y_1>0 \text{ as } y_1(1) = 5), \] \[ y_1 = e^{\frac{x}{2} - \frac{\sin 2x}{4} + C_1} = e^{C_1} e^{\frac{x}{2} - \frac{\sin 2x}{4}}. \] Apply the initial condition \( y_1(1) = 5 \): \[ y_1(1) = e^{C_1} e^{\frac{1}{2} - \frac{\sin 2(1)}{4}} = 5. \] \[ \sin 2 = \sin 2 \cdot 1 = \sin 2 \approx 0.9093, \quad \frac{\sin 2}{4} \approx 0.2273, \quad \frac{1}{2} - \frac{\sin 2}{4} \approx 0.5 - 0.2273 = 0.2727, \] \[ e^{\frac{1}{2} - \frac{\sin 2}{4}} \approx e^{0.2727} \approx 1.3135, \] \[ e^{C_1} \cdot 1.3135 = 5 \implies e^{C_1} = \frac{5}{1.3135} \approx 3.806. \] Thus: \[ y_1(x) = 3.806 e^{\frac{x}{2} - \frac{\sin 2x}{4}}. \] For the limit, we need the behavior as \( x \to 0^+ \), so approximate: \[ y_1(x) \approx 3.806 e^{\frac{x}{2}} \quad (\text{as } \sin 2x \to 0 \text{ when } x \to 0). \]
Solve for \( y_2(x) \): \[ \frac{dy_2}{dx} - (\cos x)^2 y_2 = 0 \implies \frac{dy_2}{y_2} = (\cos x)^2 \, dx. \] \[ (\cos x)^2 = \frac{1 + \cos 2x}{2}, \quad \int (\cos x)^2 \, dx = \frac{1}{2} \left( x + \frac{\sin 2x}{2} \right) + C_2 = \frac{x}{2} + \frac{\sin 2x}{4} + C_2, \] \[ \ln y_2 = \frac{x}{2} + \frac{\sin 2x}{4} + C_2, \] \[ y_2 = e^{C_2} e^{\frac{x}{2} + \frac{\sin 2x}{4}}. \] \[ y_2(1) = e^{C_2} e^{\frac{1}{2} + \frac{\sin 2}{4}} = \frac{1}{3}, \quad e^{\frac{1}{2} + \frac{\sin 2}{4}} \approx e^{0.7273} \approx 2.069, \] \[ e^{C_2} \cdot 2.069 = \frac{1}{3} \implies e^{C_2} = \frac{1}{3 \cdot 2.069} \approx 0.161, \] \[ y_2(x) \approx 0.161 e^{\frac{x}{2}} \quad (\text{as } x \to 0). \]
Solve for \( y_3(x) \): \[ \frac{dy_3}{dx} - \left( \frac{2 - x^3}{x^3} \right) y_3 = 0 \implies \frac{dy_3}{y_3} = \frac{2 - x^3}{x^3} \, dx = \left( \frac{2}{x^3} - 1 \right) dx, \] \[ \int \left( \frac{2}{x^3} - 1 \right) dx = 2 \int x^{-3} \, dx - \int dx = 2 \left( \frac{x^{-2}}{-2} \right) - x + C_3 = -\frac{1}{x^2} - x + C_3, \] \[ \ln y_3 = -\frac{1}{x^2} - x + C_3, \] \[ y_3 = e^{C_3} e^{-\frac{1}{x^2} - x}. \] \[ y_3(1) = e^{C_3} e^{-\frac{1}{1^2} - 1} = e^{C_3} e^{-2} = \frac{3}{5e}, \] \[ e^{C_3} \cdot e^{-2} = \frac{3}{5e} \implies e^{C_3} = \frac{3}{5e} \cdot e^2 = \frac{3e}{5}, \] \[ y_3(x) = \frac{3e}{5} e^{-\frac{1}{x^2} - x}. \] As \( x \to 0^+ \), \( -\frac{1}{x^2} \to -\infty \), so \( e^{-\frac{1}{x^2}} \to 0 \), and \( y_3(x) \to 0 \).
Step 2: Evaluate the limit

We need: \[ \lim_{x \to 0^+} \frac{y_1(x) y_2(x) y_3(x) + 2x}{e^{3x} \sin x}. \]
- \( y_1(x) y_2(x) y_3(x) \approx \left( 3.806 e^{\frac{x}{2}} \right) \left( 0.161 e^{\frac{x}{2}} \right) \left( \frac{3e}{5} e^{-\frac{1}{x^2} - x} \right) \),

- Constant: \( 3.806 \times 0.161 \times \frac{3e}{5} \approx 0.613 \times 1.632 \approx 1.0 \),

- Exponent: \( e^{\frac{x}{2} + \frac{x}{2} - \frac{1}{x^2} - x} = e^{-\frac{1}{x^2}} \), which dominates and \(\to 0\).

- Numerator: \( y_1 y_2 y_3 \to 0 \), so \( y_1 y_2 y_3 + 2x \approx 2x \).

- Denominator: \( e^{3x} \sin x \approx 1 \cdot x = x \),

- Limit: \[ \frac{2x}{x} = 2. \]
Final Answer: The limit is \( \boxed{2} \).