Final answer:
To verify the solution y₁(x) = e^x, we substitute it into the differential equation and simplify. The result shows that e^x satisfies the equation. To find a second linearly independent solution, we use the Reduction of Order method. We assume y₂(x) = u(x) y₁(x) and solve for u(x), ultimately obtaining y₂(x) = rac{A e^3x}{2} + B e^2x as the second solution.
Step-by-step explanation:
In order to verify that y₁(x) = e^x solves the linear homogeneous differential equation, we will substitute y₁(x) into the differential equation and check if it satisfies the equation.
Step 1: Substitute y₁(x) = e^x into the differential equation:
x y'' - (2x + 1) y' + (x + 1) y = 0
x (d²/dx²(e^x)) - (2x + 1) (d/dx(e^x)) + (x + 1) (e^x) = 0
e^x + (2x + 1) e^x + (x + 1) e^x = 0
Now simplify the equation:
(e^x)(4x + 2) = 0
Since e^x is never equal to zero, the equation (e^x)(4x + 2) = 0 is satisfied when 4x + 2 = 0. Solving for x, we get x = -1/2.
Therefore, y₁(x) = e^x is a solution to the given differential equation.
To find a second linearly independent solution, we can use the method of Reduction of Order. Let's assume y₂(x) = u(x) y₁(x), where u(x) is a function to be determined. Substitute this into the differential equation:
x (u''(x)y₁(x) + 2u'(x)y₁'(x) + u(x)y₁''(x)) - (2x + 1) (u'(x)y₁(x) + u(x)y₁'(x)) + (x + 1) (u(x)y₁(x)) = 0
Expanding and simplifying the equation gives:
u''(x) + (2 - u'(x) - u(x)) e^x = 0
To find u(x), let's assume that u(x) = v(x)e^-x. Substitute this into the above equation:
v''(x) - 2v'(x) = 0
This is a simple linear equation that can be solved by integrating twice:
v'(x) = A e^2x
v(x) = rac{A e^2x}{2} + B
Therefore, u(x) = rac{A e^2x}{2} + B e^x
Finally, the second linearly independent solution y₂(x) is given by:
y₂(x) = u(x) y₁(x) = (rac{A e^2x}{2} + B e^x)e^x = rac{A e^3x}{2} + B e^2x
The general solution of the differential equation is then:
y(x) = c₁y₁(x) + c₂y₂(x) = c₁e^x + (c₂(rac{A e^3x}{2} + B e^2x))