Question

Let \( y(x), x>0 \) be the solution of the differential equation \[ x^2 \frac{d^2y}{dx^2} + 5x \frac{dy}{dx} + 4y = 0 \] satisfying the conditions \( y(1) = 1 \) and \( y'(1) = 0 \). Then the value of \( e^2y(e) \) is .............

Hint:

For solving Cauchy-Euler equations, assume a solution of the form \( y(x) = x^r \) and solve the characteristic equation.

Answer 1

Correct Answer: 2.9 - 3.1

Step 1: Solving the differential equation.
The given differential equation is a Cauchy-Euler equation. Assuming the solution is of the form \( y(x) = x^r \), we substitute this into the equation: \[ x^2 r(r - 1) x^{r-2} + 5x r x^{r-1} + 4x^r = 0 \] This simplifies to: \[ r(r - 1) + 5r + 4 = 0 \] Solving the characteristic equation: \[ r^2 + 4r + 4 = 0 \] This gives \( r = -2 \). The general solution is: \[ y(x) = (A + B \ln(x)) x^{-2} \]
Step 2: Applying initial conditions.
Substituting the initial conditions \( y(1) = 1 \) and \( y'(1) = 0 \) into the solution, we find \( A = 1 \) and \( B = 2 \). Thus, the solution is: \[ y(x) = (1 + 2 \ln(x)) x^{-2} \]
Step 3: Finding \( e^2 y(e) \).
Substituting \( x = e \): \[ y(e) = (1 + 2 \ln(e)) e^{-2} = (1 + 2) e^{-2} = 3e^{-2} \] Thus: \[ e^2 y(e) = e^2 \cdot 3e^{-2} = 3 \]
Step 4: Conclusion.
Therefore, the value of \( e^2 y(e) \) is \( \boxed{3} \).