Answer:To solve the given second-order homogeneous differential equation, x²y" + 3xy' + y = 0, with one solution y1 = 1/x and x > 0, we can use the method of reduction of order.
Let's denote the second linearly independent solution as y2(x). We can express y2(x) as y2(x) = v(x) * y1(x), where v(x) is a function to be determined.
Now, we can substitute y2(x) into the differential equation:
x²(v'' y1 + 2v' y1'') + 3x(v' y1 + v y1') + v * y1 = 0
x²v'' + 2xv' + 3xv' + 3v + v = 0
x²v'' + 5xv' + 4v = 0
This is a first-order differential equation in v. Let z = v', then v' = z and v'' = z'. Substituting into the equation gives:
x²z' + 5xz + 4v = 0
Now, we have a first-order linear differential equation in z. We can solve this equation to find v, and then integrate v to find y2.
Finally, the general solution to the original differential equation is y(x) = C1 y1(x) + C2 y2(x), where C1 and C2 are constants determined by initial conditions or boundary conditions.
Therefore, by following these steps and solving the first-order linear differential equation in z, you can find the second linearly independent solution y2(x) to the given second-order homogeneous differential equation.
Step-by-step explanation:
