Question

Consider the ordinary differential equation \[ x^2 \frac{d^2y}{dx^2} - x \frac{dy}{dx} - 3y = 0, \] with the boundary conditions \( y(x=1) = 2 \) and \( y(x=2) = 17/2 \). The solution \( y(x) \) at \( x = 3/2 \), rounded off to 2 decimal places, is:

Hint:

For Cauchy-Euler equations, assume a solution \( y = x^r \) and solve the characteristic equation to find the roots. Use boundary conditions to solve for constants.

Answer 1

Given: \[ x^2 \frac{d^2y}{dx^2} - x \frac{dy}{dx} - 3y = 0 \quad \dots (i) \] Boundary Conditions: \[ y(x = 1) = 2, \quad y(x = 2) = \frac{17}{2}. \] This is in the form of a Cauchy linear differential equation: \[ x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0. \] Substituting \( x = e^z \): \[ x^2 \frac{d^2y}{dx^2} = D(D-1)y, \quad x \frac{dy}{dx} = Dy. \] Substituting into equation (i): \[ D(D-1)y - Dy - 3y = 0. \] \[ [D^2 - D - D - 3]y = 0. \] \[ [D^2 - 2D - 3]y = 0. \] This is in the form of a homogeneous linear differential equation: \[ [f(D)]y = 0. \] The auxiliary equation is given by: \[ f(m) = 0. \] \[ m^2 - 2m - 3 = 0. \] Solving the quadratic equation: \[ m^2 - 3m + m - 3 = 0. \] \[ m(m-3) + (m-3) = 0. \] \[ (m+1)(m-3) = 0. \] The roots are: \[ m_1 = -1, \quad m_2 = 3. \] Complementary Function (C.F.): Since the roots are real and distinct, the complementary function is: \[ C.F. = C_1 e^{m_1z} + C_2 e^{m_2z}. \] \[ C.F. = C_1 e^{-z} + C_2 e^{3z}. \] Particular Integral (P.I.): Since the equation is homogeneous, \[ P.I. = 0. \] Complete Solution: \[ y = C.F. + P.I. \] \[ y = C_1 e^{-z} + C_2 e^{3z}. \] Substituting \( z = \log x \): \[ y = C_1 e^{-\log x} + C_2 e^{3 \log x}. \] \[ y = C_1 e^{\log\left(\frac{1}{x}\right)} + C_2 e^{\log(x^3)}. \] \[ y = C_1 \frac{1}{x} + C_2 x^3 \quad \dots (ii). \] Using Boundary Conditions: 1. When \( x = 1 \), \( y = 2 \): \[ 2 = C_1 + C_2. \] 2. When \( x = 2 \), \( y = \frac{17}{2} \): \[ \frac{17}{2} = \frac{C_1}{2} + 8C_2. \] \[ 17 = C_1 + 16C_2. \] Solving for \( C_1 \) and \( C_2 \): From the first equation: \[ C_1 = 2 - C_2. \] Substitute into the second equation: \[ 17 = (2 - C_2) + 16C_2. \] \[ 17 = 2 + 15C_2. \] \[ 15C_2 = 15. \] \[ C_2 = 1, \quad C_1 = 1. \] Substituting \( C_1 \) and \( C_2 \) into equation (ii): \[ y = \frac{1}{x} + x^3. \] Calculating \( y \) at \( x = \frac{3}{2} \): \[ y = \frac{1}{x} + x^3. \] \[ y = \frac{1}{\frac{3}{2}} + \left(\frac{3}{2}\right)^3. \] \[ y = \frac{2}{3} + \frac{27}{8}. \] \[ y = 0.66 + 3.37. \] \[ y = 4.03. \] Final Answer: \[ y\left(\frac{3}{2}\right) = 4.03. \]