Kindly answer all the questions there in the attached picture. Explain how to re

Question

Kindly answer all the questions there in the attached picture.

Explain how to reduce solving an nlh degree linear differential equation to a first order linear system. Let A and B be n x n matrices. Show that [A + B)2 = A2 + 2AB + B2 if and only if A and B commute with each other. Let L(y) = f be a (nicely behaved) second order linear differential equation. I have two solutions y1, y2 to L(y) = 0, and one solution yv to L(y) = f, and that and yp(0) = 3 and y'p(0) = 2. You have one solution z to L(y) = J, iwd it satisfies z(0) = 4, z'(0) = 3. Explain why there must be constants c1 and c2 so that Hint Use to the existence and uniqueness theorem for linear differential equations the 'nicely behaved' descriptor means that all the technical conditions of that theorem are satisfied. How do we find real-valued solutions to differential equations when we initially find complex-valued solutions?

Explanation / Answer

(b) Suppose we have a linear n-th degree ODE, which is written in this form :

y(n) = a(n-1) y(n-1) + ... + a(1)y(1) + a(0) y + f(t)

Note the vector of length n : X = ( y y(1) ... y(n-1))^T

Now define the vector of size n : b = (0 .... f(t))^T and the nxn matrix :

A =

0 1 0 .... 0
0 0 1 ... 0
...
0 ......0 1
a0 .... an

Then the first order linear system equivalent is : X' = AX + b

(e)
Easy here :
(A+B)^2 = (A+B)(A+B) = A^2+AB+BA+B^2
So (A+B)^2 = A^2+2AB+B^2 <=> AB+BA = 2AB <=> AB=BA <=> A,B commutes

(h)
Define :
w = y1 + y2 + yp

w(0) = 1+0+3 = 4
w'(0) = 0+1+2 = 3

So w is a solution of L(y)=f and it satisifes w(0)=4 and w'(0)=3.
But the exercise states that we have a solution z to L(y)=f such that z(0)=4, z'(0)=3
So by the uniqueness theorem, since we assume all hypothesis are verified we have w = z = y1+y2+yp

(so what is asked is true for c1=c2=1)

(i) You use the fact that e^(ix) = cos(x)+isin(x) and you simplify the expressions, with
introduction of new constants. Let's do an example below.

For instance if the eigenvalues of a 2x2 system are 1+i and 1-i and the eigenvectors (1,1+i) and (1,1-i)
the general solution is x(t) = c1(1,1+i)^Te^((1+i)t) + c2(1,1-i)^T e^((1-i)t)
So each component of the solution is :
x1(t) = e^t ( c1e^(it)+c2e^(-it))
x2(t) = e^t ( c1(1+i)e^(it) + c2(1-i)e^(-it))

x1(t) = e^t ( cos(t)(c1+c2) + i(c1-c2)sin(t)) = e^t ( A cos(t) + Bsin(t)) with A=c1+c2, B=i(c1-c2)

x2(t) = e^t( [c1(1+i)+c2(1-i)]cos(t) + i(c1(1+i)-c2(1-i))sin(t))
x2(t) = e^t( (c1+c2 + i(c1-c2))cos(t) + (i(c1-c2)-(c1+c2))sin(t) = e^t( (A+B)cos(t) + (B-A)sin(t))

So we get the real-valued solution :

x1(t) = e^t ( Acos(t) + Bsin(t))
x2(t) = e^t ( (A+B)cos(t)+(B-A)sin(t))

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