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 xᵃ⁺ᶦᵇ = ⁽ᵃ⁺ᶦᵇ⁾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

To find a basis for ker(L), assume y = xr and solve for r. Calculate the Wronskian of the solution. Solve the initial value problem using undetermined coefficients.

Step-by-step explanation:

To find a basis for ker(L), we assume that there exists a complex number, r, such that y = xr ε ker(L). We can substitute y = xr into the equation L[y] = x² y’’ + 5 x y’ + 8y and solve for r. By separating the solution into real and imaginary parts, we can find the values of r that satisfy the equation. Once we have the values of r, we can construct the basis for ker(L) by taking the corresponding values of x and y.

To calculate the Wronskian of the solution obtained in part A, we need to evaluate the determinant of the matrix formed by the real and imaginary parts of the solution. The Wronskian is given by W(y₁, y₂) = -2x³ or W(y₁, y₂) = 2x³, depending on how the columns are lined up.

To solve the initial value problem L[y] = x² with y(1) = y’(1) = 0, we can use the method of undetermined coefficients. Since L is a polynomial differential operator of degree 2, we can assume a particular solution of the form y = Ax³ + Bx² + Cx + D. Substituting this into the equation and solving for the coefficients will give us the particular solution. By adding the particular solution to the general solution obtained in part A, we can find the complete solution to the initial value problem.