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course Phy 279
3/6 12
q_a_10________________________________________
See also Class Notes 110223
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Question:
`q001. Using the substitution y = A e^(r t) find two linearly independent solutions to the equation y '' + y ' - 6 y = 0.
y = A e^(r t) for y '' + y ' - 6 y = 0.
Ar^2e^(rt) + Are^(rt) - 6 = 0
r^2 + r - 6 = 0
r = 2 and r = -3
solutions are
e^(2t) and e^(-3t)
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Show that your solutions are linearly independent.
e^(2t) e^(-3t)
2e^(2t) -3 e^(-3t)
= f(t)*g’(t) - g(t)*f’(t)
= e^(2 t)*-3e^(-3t) - e^(-3 t)*2e^(2t)
= e^(2 t)*-3e^(-3t) - e^(-3 t)*2e^(2t) = -5e^(-t)
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Form two different linear combinations of your solutions and show whether they are linearly independent.
I’m not sure how to answer this question but would we just calculate the determinates of e^(2t) and e^(-3t) with different scalar multiples applied?????
@& I would choose nonzero coefficients to make the linear combinations.
For example
y_1 = 2 e^(2 t) + 5 e^(-3 t)
and
y_2 = 3 e^(2 t) - 2 e^(-3 t).*@
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03-07-2011