Let y be a solution to the differential equation
\[ y'' - 4xy' + (4x^2 - 1)y = -3e^{x^2} \sin(2x) \],
with boundary conditions \( y(0) = 0 \) and \( y(\frac{\pi}{2}) = 0 \).
Then the value of \( y(2\pi) \) is:
(a) \( \frac{e^{\pi^2}}{4} \)
(b) 0
(c) \( e^{4\pi^2} \)
(d) 1
Answer (1)
To solve the differential equation y ′′ − 4 x y ′ + ( 4 x 2 − 1 ) y = − 3 e x 2 sin ( 2 x ) with the given boundary conditions y ( 0 ) = 0 and y ( 2 π ) = 0 , we aim to find the value of y ( 2 π ) .
Analyzing the Problem
This is a second-order, linear, non-homogeneous differential equation with variable coefficients. It's typically tackled using methods such as:
Variation of Parameters or Undetermined Coefficients for the particular solution.
Solving the homogeneous equation y ′′ − 4 x y ′ + ( 4 x 2 − 1 ) y = 0 .
Homogeneous Equation
The homogeneous part y ′′ − 4 x y ′ + ( 4 x 2 − 1 ) y = 0 is a type of equation that often involves special functions like Hermite polynomials, but in this context, we return our focus to the original differential problem.
Particular Solution
For the non-homogeneous part, the function on the right side is − 3 e x 2 sin ( 2 x ) . This suggests the method of undetermined coefficients is impractical due to the complex nature of the forcing function. Thus, use of variation of parameters is more suitable, albeit involved.
Applying the Boundary Conditions
With the boundary conditions y ( 0 ) = 0 and y ( 2 π ) = 0 , any particular solution satisfying these will do so over the interval. Without the explicit form of y ( x ) , and given the complexity, solving the boundary value problem analytically can often be assisted by numerical methods such as the finite difference method.
Checking Possible Values
Given that the boundary conditions and the differential form are such that they imply symmetry or periodic behavior, particularly within the concern of eigenvalues, it's plausible that the solution at y ( 2 π ) remains zero as well due to its structural nature and boundary conditions:
Thus, evaluating symmetrical boundary control and influence across the entire boundary [ 0 , 2 π ] leads us to conclude that the inherent periodicity detailed in the setup hints at:
(b) 0
Therefore, the value of y ( 2 π ) is most reasonably (b) 0 .
