Answer:
Explanation:
To find the special solution of the given differential equation using the method of changing parameters, we assume the solution in the form:
y = c1(x)y1(x) + c2(x)y2(x)
where y1(x) and y2(x) are the fundamental solutions of the homogeneous equation y'' + 2y' + y = 0.
The correct step for the process is:
C) c2'(x) = √(x+1)
The reason is that when we differentiate the equation y = c1(x)y1(x) + c2(x)y2(x) with respect to x, we get:
y' = c1'(x)y1(x) + c1(x)y1'(x) + c2'(x)y2(x) + c2(x)y2'(x)
For the special solution, we want the term c1(x)y1(x) + c2(x)y2(x) to satisfy the original non-homogeneous equation y'' + 2y' + y = 3e^(-x)√(x+1). Therefore, we need the terms involving the derivatives of c1(x) and c2(x) to cancel out the terms in the non-homogeneous equation.
In this case, we have:
y' = c1'(x)y1(x) + c2'(x)y2(x)
Comparing this with the derivative term in the non-homogeneous equation, which is 3e^(-x)√(x+1), we can see that the correct step is to set c2'(x) = √(x+1).
Therefore, the answer is C) c2'(x) = √(x+1).