Question

If a function \(y(x)\) is described by the initial-value problem, \(\dfrac{d^2y}{dx^2} + 5\dfrac{dy}{dx} + 6y = 0\), with initial conditions \(y(0) = 2\) and \(\dfrac{dy}{dx}\bigg|_{x=0} = 0\), then the value of \(y\) at \(x = 1\) is ................. . (Round off to 2 decimal places)

Hint:

For homogeneous linear ODEs with constant coefficients, always find characteristic roots and use initial conditions to solve for constants.

Answer 1

Correct Answer: 0.6

Step 1: Write the characteristic equation.
The differential equation is: \[ \frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 6y = 0 \] Characteristic equation: \[ r^2 + 5r + 6 = 0 $\Rightarrow$ (r + 2)(r + 3) = 0 \] \[ r = -2, -3 \]

Step 2: General solution.
\[ y(x) = A e^{-2x} + B e^{-3x} \]

Step 3: Apply initial conditions.
For \(x = 0,\ y(0) = 2:\) \[ A + B = 2 (1) \] Derivative: \[ \frac{dy}{dx} = -2A e^{-2x} - 3B e^{-3x} \] At \(x=0,\ \frac{dy}{dx} = 0:\) \[ -2A - 3B = 0 $\Rightarrow$ 2A + 3B = 0 (2) \] Solving (1) and (2): Multiply (1) by 2 → \(2A + 2B = 4\) Subtract from (2): \[ (2A + 3B) - (2A + 2B) = 0 - 4 $\Rightarrow$ B = -4 \] Substitute in (1): \[ A - 4 = 2 $\Rightarrow$ A = 6 \]

Step 4: Substitute constants.
\[ y(x) = 6 e^{-2x} - 4 e^{-3x} \] At \(x = 1:\) \[ y(1) = 6 e^{-2} - 4 e^{-3} = 6(0.1353) - 4(0.0498) = 0.8118 - 0.1992 = 0.6126 \] Rounded to two decimals: \(y(1) = 0.61\).

Step 5: Conclusion.
Hence, \(y(1) = 0.61\).