Final answer:
By calculating the first and second derivatives of y₁(x)=e and substituting them into the given differential equation, it can be shown that the terms cancel out, verifying that y₁(x)=e is indeed a solution to the equation (x-1)y'' - xy' + y = 0.
Step-by-step explanation:
We are asked to verify that y₁(x)=e is a solution to the differential equation (x-1)y'' - xy' + y = 0. To do this, we need to find the first and second derivatives of y₁(x), substitute them into the equation, and check if the equation holds true.
First, we compute the first derivative of y₁(x):
y'₁(x) = d/dx(e) = 3e.
Next, we compute the second derivative of y₁(x):
y''₁(x) = d/dx(3e) = 9e.
Now we substitute y₁, y'₁, and y''₁ into the original equation:
(x - 1)(9e) - x(3e) + e = 0
Expanding and simplifying gives us:
(9xe - 9e) - (3xe) + e = 9xe - 9e - 3xe + e = 0
Clearly, the terms on the left side cancel each other out resulting in 0, which shows that y₁(x)=e indeed satisfies the differential equation.