Solutions To Linear Differential Equations Problems

1) Obtain the general solution(s) for the system:

dx/dt = 2x + y

 dy/dt = 2x + 3y

Solution:

 In this case, the alert math maven will note the x, y derivatives are with respect to t, but the equivalence is to x, y forms with no t variable present. We reconcile our terms and allow the solution to proceed via the following substitutions:

x = A exp(lt) and y = B exp(lt)

With these substitutions we than re-write our system in a consistent form:

l A exp(lt)   =  2 A exp(lt)    + B exp(lt)

lB exp(lt)   =  2 A exp(lt)    + 3B exp(lt)

(Recall: dx/dt = d (A exp(lt))/dt = l A exp(lt) - Analogous for dy/dt!)

Now, collect like terms:

2 A exp(lt) - l A exp(lt)       + B exp(lt) =  0

2 A exp(lt)    + 3B exp(lt)   -lB exp(lt)   = 0

Factor out the exp(lt) terms top and bottom to get:

(2 - l) A  + B = 0

2A  + (3 - l) B = 0

Set up the determinant as per prior problems: (A - l) D = 0 =

(2 - l…………1)
(2………...3 - l )

Find the characteristic equation, viz.

(2 - l) (3 - l) – 2 = 0 = 6 - 3 l - 2 l + l² – 2 =

l² – 5l  + 4 = 0

Or: (l - 4) (l - 1) = 0

So we have eigenvalues : l1 = 4  and l2 = 1

As before with previous problems, substitute the eigenvalue l1 = 4   into the matrix to get:

[-2 ……..1] [A]

[2……..-1] [B]


Whence: -2 A + B = 0 and 2A – B = 0


So, A = 1, B = 2, so c1 =

[1]
[2]

The solution is then: X1 = c1 exp (4t)

   

Now, substitute the eigenvalue l2 = 1   into the matrix to get:

  [1 ……..1] [A]

  [2……..2] [B]

Whence:  A + B = 0 and 2A + 2B = 0 so: A = 1 and B = -1

so c2 =

[1]
[-1]

The solution is then: X2 = c2 exp (t)

The full general solution for the system is:

X = X1 + X2 = c1 exp (4t) +  c2 exp (t)  

2)Obtain the general solution(s) for the system:

dx1/ dt = 3x1 - 4x2

dx2/dt = x1 - x2

 When roots are real and equal, then assume soln. of form:

x1 = A exp(lt) and x2 = B exp(lt)

If  Det (A - lI) D = 0 only has one soln.  l  then (A - lI) D = 0 has at least one independent solution.  A first solution is:  X1 =  D exp (lt)

Now take:  X =  D exp (lt)  + Bt exp (lt)

dX/ dt =  l D exp (lt)  +  B exp(lt) +  l Bt exp(lt)

=  AX  +  AD exp (lt)  + ABt exp (lt)

For all t we need to satisfy:

B  =   (A - lI) D  +   t (A - lI) B

Take  (A - lI) B  =  0

 (A - lI) D  =  B

The independent solution is then:

X2 =   D exp (lt)  + Bt exp (lt)

Applied to the given linear system we get:

And:

DET

The characteristic equation is then:

 l² – 2l  + 1 = 0

Then write:   (A - lI) D  = 

(2.........-4) (d1 )       (0)

(1..........-2) (d2) =   (0)

Whence:   

2d1 – 4d2 = 0

 d1  -  2d2  = 0

 =>  3d1 - 6d2 = 0

Or:  d1 - 2d2 = 0

Then:

d1 = 2d2   so:   d1 = 2  and d2 = 1

Eigenvector = 

(2)

(1)

Finally: