Question

Let L[y] = x2 y’’ + 5 x y’ + 8y.

A) By assuming that there exists a complex number, r, such that y
= xr ε ker(L)
find a basis for ker(L). You should show all the steps – don’t just plug into a formula. You will need to use
the fact that x+ = (+)ln (x) and separate the solution into real and imaginary parts.
B) Calculate the
Wronskian of your solution from A. You should get either W(y1, y2) = 2x3 or
W(y1, y2) = -2x3 , depending on how you line up your columns.
C) Solve the initial value problem: L[y] = x2 , y(1) = y’(1) = 0. You may use reduction of order or variation
of parameters to find a particular solution, or – much easier – notice that L : P2 -> P2 , so you can use the
method of undetermined coefficients.

Answer 1

This answer provides step-by-step instructions for finding a basis for the kernel of the given differential operator L, calculating the Wronskian of the basis vectors, and solving the initial value problem L[y] = x^2 with specific initial conditions.

Step-by-step explanation:

Part A) Finding a basis for ker(L)

To find a basis for ker(L), we need to find complex numbers r such that y = xr satisfies L[y] = 0. Plugging in the given expression for L[y] and substituting y = xr, we get a differential equation in x and r. By separating the real and imaginary parts, we can find the values of r and the corresponding basis vectors for ker(L).

Part B) Calculating the Wronskian

To calculate the Wronskian of the basis vectors found in part A, we can evaluate their derivatives and use them to construct a matrix. Taking the determinant of this matrix gives us the Wronskian, which may be either 2x^3 or -2x^3, depending on how the columns are arranged.

Part C) Solving the initial value problem

To solve the initial value problem L[y] = x^2 with y(1) = y'(1) = 0, we can use the method of undetermined coefficients. Since L is a linear operator from P2 to P2 (the space of degree 2 polynomials), we can find a particular solution by equating the coefficients of the polynomial x^2. We can then find the general solution by adding the complementary solution obtained from parts A and B to the particular solution.