Question

Let \( y(x) \) be the solution of the initial value problem, \[ x^2 y'' + xy' - y = 0, x>0, y(1) = 0, y'(1) = 2. \] Then the value of \( y'\left(\frac{1}{2}\right) \) is equal to (Answer in integer) ................

Hint:

Cauchy-Euler equations can often be solved using trial solutions of the form \( y = x^r \). The characteristic equation helps determine the powers involved.

Answer 1

We are given a second-order linear differential equation with variable coefficients:
\[ x^2 y'' + x y' - y = 0 \]
This is a Cauchy-Euler equation. Let us try a solution of the form \( y = x^r \).
Then:
\[ y' = r x^{r-1}, y'' = r(r-1)x^{r-2} \]
Substitute into the differential equation:
\[ x^2 \cdot r(r-1)x^{r-2} + x \cdot r x^{r-1} - x^r = r(r-1)x^r + r x^r - x^r = 0 \]
\[ \Rightarrow \left[ r(r-1) + r - 1 \right] x^r = 0 \Rightarrow r^2 - 1 = 0 \Rightarrow r = \pm1 \]
So the general solution is:
\[ y(x) = A x + B x^{-1} \]
Now apply initial conditions:
Condition 1: \( y(1) = 0 \Rightarrow A(1) + B(1^{-1}) = A + B = 0 \Rightarrow A = -B \)
Condition 2: \( y'(x) = A - B x^{-2} \Rightarrow y'(1) = A - B = 2 \)
Substitute \( A = -B \) into this:
\[ -B - B = 2 \Rightarrow -2B = 2 \Rightarrow B = -1 \Rightarrow A = 1 \]
So the solution is: \[ y(x) = x - \frac{1}{x} \]
Then, \[ y'(x) = 1 + \frac{1}{x^2} \]
Thus, \[ y'\left(\frac{1}{2}\right) = 1 + \left(\frac{1}{(1/2)^2}\right) = 1 + 4 = 5 \]