Find the Wronskian (W) of the fundamental solution of the following differential equation:
y'' - 2y' + 1 = (x+1)e^{2x}
A. W = e^x
B. W = e^{2x}
C. W = xe^{2x}
D. W = x
Answer (2)
The Wronskian of the fundamental solutions of the differential equation y ′′ − 2 y ′ + 1 = ( x + 1 ) e 2 x is found by first solving the homogeneous equation, resulting in the fundamental solutions e x and x e x . The Wronskian is then calculated to be e 2 x . Thus, the correct option is B. W = e^{2x} .
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The Wronskian is used to determine whether a set of solutions is linearly independent for a given differential equation. For the equation given, y ′′ − 2 y ′ + 1 = ( x + 1 ) e 2 x , we need to focus on the homogeneous part first:
y ′′ − 2 y ′ + y = 0
To find the fundamental solutions of this homogeneous equation, we solve the characteristic equation derived from the differential equation:
r 2 − 2 r + 1 = 0
Solving the quadratic equation gives us:
( r − 1 ) 2 = 0
This indicates a repeated root r = 1 . Therefore, the fundamental solutions of the homogeneous equation are:
y 1 = e x , y 2 = x e x
To compute the Wronskian W ( y 1 , y 2 ) of these solutions, we use the formula for the Wronskian of two functions f and g :
W ( y 1 , y 2 ) = y 1 y 1 ′ y 2 y 2 ′
Substituting our solutions:
y 1 = e x , y 2 = x e x
y 1 ′ = e x , y 2 ′ = e x + x e x
Calculate the determinant:
W ( y 1 , y 2 ) = e x e x x e x ( 1 + x ) e x = e x [( 1 + x ) e x ] − x e x [ e x ]
Simplify:
e 2 x + x e 2 x − x e 2 x = e 2 x
Thus, the Wronskian is W = e 2 x , which corresponds to option (B).
