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course Mth 279
April 12 around 12:15pm.
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Question:
`q001. Consider the equation y '' + 4 y ' - 5 y = 0. Substitute y = e^(r t) and solve for r.
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Y = e^(rt), y’ = Are^(rt), y” = Ar^2e^(rt)
(Ar^2e^rt) + 4(Are^rt) - 5(Ae^rt) = 0
r^2 + 4r -5 = 0
(r+5)(r-1) = 0
R = -5, 1
Any linear combination of {e^t, e^-5t} is a solution.
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You should have obtained two values of r. For the moment call them r_1 and r_2.
Show that the following functions solve the above differential equation, for your values of r:
y = A e^(r_1 * t), where A can be any constant
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R1 = -5
R2 = 1
Y = Ae^(-5t)
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y = B e^(r_2 * t), where B can be any constant
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Y = Be^(t)
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y = A e^(r_1 * t) + B e^(r_2 * t), where A and B can be any constants.
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Y = Ae^(-5t) + Be^(t)
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If we want a solution with the conditions y(0) = 1 and y ' (0) = -1, can we adjust the constants A and B so that y = A e^(r_1 * t) + B e^(r_2 * t) is a solution?
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Well, couldn’t you verify this by linear algebra?!
W(e^t, e^-5t) = determinant matrix and see if the det = 0?!
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It is shown in the Class Notes that we obtain the equation r^2 + 4 r - 5, with solutions r = 1 and r = -5.
Using r_1 = 1 and r_2 = 5 our solutions are y = e^t and y = e^(-5 t).
If we substitute y = A e^t into our equation we get
(A e^t) '' + 4 (A e^t) ' - 5 A e^t = 0
so that
A e^t 4 A e^t - 5 A e^t = 0,
which is clearly so.
If we substitute y = B e^(-5 t) we get
(B e^(-5 t)) '' + 4 (B e^(-5 t)) ' - 5 B e^(-5 t) = 0
so that
25 (B e^(-5 t)) - 20 (B e^(-5 t)) ' - 5 B e^(-5 t) = 0,
again clearly true.
If we substitute A e^t + B e^(-5 t) we get
(A e^t + B e^(-5 t)) '' + 4 (A e^t + B e^(-5 t)) ' - 5 (A e^t + B e^(-5 t)) = 0
so that
(A e^t) ''+ (B e^(-5 t)) '' + 4 (A e^t) ' + 4 (B e^(-5 t)) '- 5 A e^t - 5 B e^(-5 t) = 0
which we easily rearrange to
(A e^t) '' + 4 (A e^t) ' - 5 A e^t + (B e^(-5 t)) '' + 4 (B e^(-5 t)) ' - 5 B e^(-5 t) = 0
which becomes
A e^t 4 A e^t - 5 A e^t + 25 (B e^(-5 t)) - 20 (B e^(-5 t)) ' - 5 B e^(-5 t) = 0,
which is easily verified.
If we want y(0) = 1 we have to satisfy
y(0) = A e^0 + B e^(-5 * 0)) = 0
so that
A + B = 1,
and if we also have the condition y ' (0) = -1 we have
y ' = (A e^t + B e^(-5 t)) ' = A e^t - 5 B e^(-5 t) so that
y ' (0) = A e^0 - 5 B e^(-5 * 0) = 0
A - 5 B = -1.
The two simultaneous equations
A + B = 1
A - 5 B = -1
are easily solved. We obtain
A = 2/3, B = 1/3
so our the solution satisfying the initial conditions y(0) = 1, y ' (0) = -1 is
y = 2/3 e^t + 1/3 e^(-5 t).
As shown in the Class Notes, the Wronskian of this solution set is
W(t) = det [ y_1 y_2; y_1 ' y_2 ' ]
= det [ e^t , e^(-5 t); (e^t) ' , e^(-5 t ) ' ]
= det [ e^t , e^(-5 t); (e^t) , -5e^(-5 t ) ]
= -6 e^(-4 t).
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Question:
`q002. Suppose f(t) = e^(2 t) and g(t) = e^(-3 t).
Show that the determinant of the 2 x 2 matrix [ f, g ; f ', g ' ] (that is the matrix with first row f g and second row f ' g ' ) is not identically zero.
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I used linear algebra and found the det to see if it was 0 or not.
After solving, I got:
=[-5e^(-t)] doesn’t equal 0
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Suppose f(t) = t^2 and g(t) = t^3.
Show that the determinant of the 2 x 2 matrix [ f, g; f ', g ' ] is not identically zero.
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I used linear algebra and found the det to see if it was 0 or not.
After solving, I got:
=[t^4] doesn’t equal 0
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Show that if f is a constant multiple of g, then the determinant of the 2 x 2 matrix [ f, g; f ', g ' ] is identically zero.
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Being linearly independent, you do the det and derive and see if it cancels out. If it doesn’t equal zero, then it’s linearly independent. If all cancels out, NOT linearly independent.
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If f is a constant multiple of g then we can write f = c * g for some constant c, so that our determinant is
det [ c * g , g ; (c * g) ' , g ' ]
= det [ c * g , g ; c * g ' , g ' ]
= (c * g) * g ' - g * (c * g ' )
= c g g ' - c g g ' = 0.
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Question:
`q003. Recall that the span of two vectors v_1 and v_2 is the set of all their possible linear combinations v = a_1 v_1 + a_2 v_2, where a_1 and a_2 are scalar quantities.
Recall also that a set of vectors is linearly independent if there is no nontrivial linear combination of those vectors which is equal to zero, and that this condition is equivalent to saying that none of the vectors can be expressed as a linear combination of the others.
Functions can be regarded as vectors, with the scalars being the real numbers. Two functions are said to be linearly independent if neither is a constant multiple of the other.
If the functions are f and g, then this condition is equivalent to saying that the determinant of [ f, g; f ', g ' ] is not identically zero.
In this case the set of linear combinations A f + B g is said to span a function space of two dimensions.
Show that the two functions f(t) = sin(t) and g(t) = cos(t) are linearly independent.
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Doing the det, I get:
= -sin(t^2) - cos(t^2)
Would I derive this next or leave it as is, to determine if linearly independent?!
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Show that the function 2 sin(t) + 4 cos(t) is in the span of the set {f, g}.
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Do you find the span from the general solution?!
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Show that the function t sin(t) - 2 cos(t) is not in the span of the set {f, g}.
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Show that the set {e^t, e^(-t) } is linearly independent.
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If you do the det, I got:
= (-2)
It doesn’t equal 0, therefore, linearly independent.
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Show that any function in the span of the set {e^t, e^(-t)} is a solution of the differential equation y '' - y = 0.
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Y = Ae^(rt)
Y’ = Are^(rt)
Y’’ = Ar^2e^(rt)
Would I solve for “r” and see what the general solution is?!
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W(t) is never zero so the two functions are linearly independent for all values of t.
A function in the span of the set { e^t, e^(-t) } is of the form y(t) =c_1 e^t + c_2 e^(-t), with derivative y ' = c_1 e^t - c_2 e^(-t) and second derivative y '' = c_1 e^t + c_2 e^(-t).
Thus
y '' - y = c_1 e^t + c_2 e^(-t) - ( c_1 e^t + c_2 e^(-t) ) = 0.
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Question:
`q004. Show that e^((i + 2) * t) and e^( (2 - i) * t) are linearly independent.
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Doing the det:
= (e^((i+2)t))*((2-i)*e^((2-i)t)) - (e^((2-i)t))*((i+2)*e^((i+2)t))
I’m not sure what to do after this or the next question…
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If c_1 = a_1 + i * b_1 and c_2 = a_2 + i * b_2, then what is the general form of c_1 e^( (i + 2) * t) + c_2 e^( (2 - i) * t), in terms of sines and cosines? You get something pretty messy, so simplify it as much as you can.
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The Taylor expansion for e^x is 1 + x + x^2 / 2 ! + x^3 / 3 ! + x^4 / 4 ! + ...
The Taylor expansion for sin(x) is x - x^3 / 3 ! + x^5 / 5 ! + ...
The Taylor expansion for cos(x) is 1 - x^2 / 2 ! + x^4 / 4 ! + ...
Substituting i * theta for x in the Taylor expansion of e^x we get
e^(i * theta) = 1 + (i * theta) + (i * theta)^2 / 2 ! + (i * theta)^3 / 3 ! + (i * theta)^4 / 4 ! + ...
= 1 + (i * theta) - theta^2 / 2 ! - i * theta ^3 / 3 ! + theta^4 / 4 ! + ...
= ( 1 - theta^2 / 2 ! + theta^4 / 4 ! + ... ) 1 + i * ( theta - theta ^3 / 3 ! ... )
= cos(theta) + i sin(theta).
The derivation of the formulas for cos(alpha + beta) and sin(alpha + beta) is shown in the Class Notes.
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&#Your work looks good. See my notes. Let me know if you have any questions. &#