Question

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

Hint:

Cauchy–Euler ODEs can be solved quickly by trial solution $y=x^m$. Always check roots of the auxiliary equation and then apply initial conditions.

Answer 1

Correct Answer: 5

Step 1: Recognize type of equation.
The equation is a Cauchy–Euler equation: \[ x^{2}y'' + xy' - y = 0. \]

Step 2: Assume trial solution.
Let $y = x^{m}$. Then: \[ y' = m x^{m-1}, y'' = m(m-1)x^{m-2}. \] Substitute: \[ x^{2}y'' + xy' - y = m(m-1)x^{m} + mx^{m} - x^{m} = (m^{2}-1)x^{m}. \] So, $m^{2}-1=0 \;\Rightarrow\; m=\pm 1$.

Step 3: General solution.
Thus, \[ y(x) = C_{1}x + C_{2}x^{-1}. \]

Step 4: Apply initial conditions.
At $x=1$: \[y(1)=C_{1}\cdot 1 + C_{2}\cdot 1 = C_{1}+C_{2}=0 \;\Rightarrow\; C_{2}=-C_{1}.\] Derivative: \[ y'(x) = C_{1} - C_{2}x^{-2}.\] At $x=1$: \[y'(1)=C_{1}-C_{2}=2.\] Since $C_{2}=-C_{1}$, we get $y'(1)=C_{1}-(-C_{1})=2C_{1}=2$, so $C_{1}=1, C_{2}=-1$.



Step 5: Explicit solution.
\[ y(x)=x - \frac{1}{x}. \] Then \[ y'(x)=1+\frac{1}{x^{2}}. \] At $x=\tfrac{1}{2}$: \[ y'\!\left(\tfrac{1}{2}\right)=1+\left(\tfrac{1}{(1/2)^{2}}\right)=1+4=5. \] Final Answer: $\boxed{5}$