1. The operator point of view. Let a be a non-zero real number, and consider the

Question

1.       The operator point of view. Let a be a non-zero real number, and consider the differential equation y

The operator point of view. Let a be a non-zero real number, and consider the differential equation y'' - 2ay' +a2y = 0. Find the differential operator L for which the above equation is the homogeneous equation L(y)=0. Show that L can be written as the composition of two first order operators, L=(d/dx - a) ? (d/dx - a) (recall that two functions, f and g, are the same if they agree on all values: f(x)=g(x) for all x?D, where D is the common domain. Then, since an operator is itself a function, two operators, L and P, are the same if they agree on all values in their domain, L(f) = P(f).) Solve the first order equation (d/dx - a)y = 0 By the result of part (c), we have (d/dx - a)(Ceax) = 0. Now use linearity to argue that L(Ceax) = 0. Also verify directly that this is so. Show that if y solves (d/dx - a)y= Ce^(ax), Then L(y) = 0 Find y solving (d/dx - a)y= Ce^(ax) Find the general solution to the original differential equation. Make sure you understand how the reasoning developed here leads to this solution Series solutions. First solve the differential equation y'-y=e^(x) by earlier methods. Then, use series methods to solve the differential equation y'-y=e^(x). work on your series solution until you can show that this solution agrees with the solution obtained by earlier methods.

Explanation / Answer

1)

y'' - 2ay' + a^2 y = 0

a)

let d/dx = D

we get

d^2/dx^2 = D^2

so,

y'' - 2ay' + a^2 y = 0

D^2 y - 2a D y + a^2 y = 0

=>

[D^2 - 2aD + a^2] y = 0

so,

differential operator L = D^2 - 2aD + a^2

b)

L = D^2 - 2aD + a^2

   = D^2 - aD - aD + a^2

= D(D-a) - a(D-a)

= (D-a)(D-a)

so,

L = (D-a) o (D-a)

c)

(d/dx

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