Question

Consider the second-order linear ordinary differential equation \[ x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} - y = 0, x \ge 1 \] with the initial conditions \[ y(x = 1) = 6, \frac{dy}{dx}\bigg|_{x=1} = 2. \] The value of y at x = 2 equals ................ (Answer in integer)

Hint:

For a second-order Cauchy-Euler equation of the form \(ax^2y'' + bxy' + cy = 0\), the auxiliary equation is always \(am(m-1) + bm + c = 0\). Recognizing this pattern can save you the time of re-deriving it during an exam.

Answer 1

Step 1: Understanding the Concept:
The given differential equation is a Cauchy-Euler equation, which has the general form \(a_n x^n y^{(n)} + \dots + a_1 x y' + a_0 y = 0\). These equations can be solved by assuming a solution of the form \(y = x^m\).
Step 2: Key Formula or Approach:
1. Assume a solution \(y = x^m\).
2. Find the first and second derivatives: \(y' = mx^{m-1}\) and \(y'' = m(m-1)x^{m-2}\). 3. Substitute these into the differential equation to find the auxiliary (or characteristic) equation for \(m\).
4. Solve the auxiliary equation to find the roots \(m_1\) and \(m_2\).
5. Write the general solution based on the roots.
6. Use the initial conditions to find the values of the constants in the general solution. 7. Evaluate the final solution at \(x=2\).
Step 3: Detailed Calculation:
1. Find the auxiliary equation:
Substitute \(y = x^m\), \(y' = mx^{m-1}\), \(y'' = m(m-1)x^{m-2}\) into the ODE: \[ x^2 [m(m-1)x^{m-2}] + x [mx^{m-1}] - [x^m] = 0 \] \[ m(m-1)x^m + mx^m - x^m = 0 \] Divide by \(x^m\) (since \(x \ge 1\), \(x^m \neq 0\)): \[ m(m-1) + m - 1 = 0 \] \[ m^2 - m + m - 1 = 0 \] \[ m^2 - 1 = 0 \] 2. Solve for m: \[ m^2 = 1 \implies m = \pm 1 \] The roots are \(m_1 = 1\) and \(m_2 = -1\).
3. Write the general solution:
Since the roots are real and distinct, the general solution is: \[ y(x) = C_1 x^{m_1} + C_2 x^{m_2} = C_1 x^1 + C_2 x^{-1} = C_1 x + \frac{C_2}{x} \] 4. Apply initial conditions:
- First condition: \(y(1) = 6\) \[ 6 = C_1 (1) + \frac{C_2}{1} \implies C_1 + C_2 = 6 \text{(Eq. 1)} \] - Second condition: \(y'(1) = 2\). First, find the derivative of the general solution: \[ y'(x) = C_1 - \frac{C_2}{x^2} \] Now apply the condition at \(x=1\): \[ 2 = C_1 - \frac{C_2}{1^2} \implies C_1 - C_2 = 2 \text{(Eq. 2)} \] 5. Solve for \(C_1\) and \(C_2\):
Add (Eq. 1) and (Eq. 2): \[ (C_1 + C_2) + (C_1 - C_2) = 6 + 2 \] \[ 2C_1 = 8 \implies C_1 = 4 \] Substitute \(C_1 = 4\) into (Eq. 1): \[ 4 + C_2 = 6 \implies C_2 = 2 \] 6. Write the particular solution: \[ y(x) = 4x + \frac{2}{x} \] 7. Evaluate at x=2: \[ y(2) = 4(2) + \frac{2}{2} = 8 + 1 = 9 \] Step 4: Final Answer:
The value of y at x = 2 is 9.
Step 5: Why This is Correct:
The problem correctly identifies the ODE as a Cauchy-Euler type and applies the standard solution method. The auxiliary equation is correctly derived and solved, and the initial conditions are used to find the specific solution, which is then evaluated at the required point.