Question

Consider the ordinary differential equation \[ x^2 \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + 2y = 0 \] with \( y(x) \) as a general solution. Given the values of \( y(1) = 1 \), \( y(2) = 5 \), the value of \( y(3) \) is equal to ______

• A. 9
• B. 12
• C. 15
• D. -15

Hint:

For Cauchy-Euler equations, try solutions of the form \( y = x^m \), which reduce the differential equation to an algebraic equation in \( m \).

Answer 1

The Correct Option is B

We are given a Cauchy-Euler (or equidimensional) differential equation:
\[ x^2 y'' - 2x y' + 2y = 0 \]
Let \( y = x^m \Rightarrow y' = m x^{m-1},\ y'' = m(m-1)x^{m-2} \)
Substitute in the equation:
\[ x^2 \cdot m(m-1)x^{m-2} - 2x \cdot m x^{m-1} + 2 x^m = 0 \]
\[ m(m-1)x^m - 2m x^m + 2x^m = 0 \]
\[ x^m \left( m(m-1) - 2m + 2 \right) = 0 \]
\[ x^m \left( m^2 - 3m + 2 \right) = 0 \Rightarrow m^2 - 3m + 2 = 0 \]
Solving the quadratic:
\[ m = 1,\ 2 \Rightarrow \text{Solutions are linearly independent: } y(x) = A x + B x^2 \]
Now apply initial conditions:
Given \( y(1) = 1 \Rightarrow A(1) + B(1)^2 = A + B = 1 \) \hfill (1)
Given \( y(2) = 5 \Rightarrow A(2) + B(4) = 2A + 4B = 5 \) \hfill (2)
From (1): \( A = 1 - B \)
Substitute into (2):
\[ 2(1 - B) + 4B = 5 \Rightarrow 2 - 2B + 4B = 5 \Rightarrow 2B = 3 \Rightarrow B = \frac{3}{2} \]
Then \( A = 1 - \frac{3}{2} = -\frac{1}{2} \)
So the solution is \( y(x) = -\frac{1}{2}x + \frac{3}{2}x^2 \)
Now compute \( y(3) \):
\[ y(3) = -\frac{1}{2}(3) + \frac{3}{2}(9) = -\frac{3}{2} + \frac{27}{2} = \frac{24}{2} = 12 \]