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Unformatted text preview: = y0 y (0) = 2C1 e3 + 2C2 e−3 = v0
• Solve this system for C1, C2... Independence and the Wronskian (Section 3.2)
y1 (t) = e2t+3 and y2 (t) = e2t−3 are two
• Example: Suppose
solutions to some equation. Can we solve ANY initial
y
condition . (0) = y0 , y (0) = v0 with these two solutions?
. y (t) = C1 e2t+3 + C2 e2t−3
y (0) = C1 e3 + C2 e−3 = y0 y (0) = 2C1 e3 + 2C2 e−3 = v0
• Solve this system for C1, C2...
• Can’t do it. Why? Independence and the Wronskian (Section 3.2)
y1 (t) = e2t+3 and y2 (t) = e2t−3 are two
• Example: Suppose
solutions to some equation. Can we solve ANY initial
y
condition . (0) = y0 , y (0) = v0 with these two solutions?
. y (t) = C1 e2t+3 + C2 e2t−3
y (0) = C1 e3 + C2 e−3 = y0 y (0) = 2C1 e3 + 2C2 e−3 = v0
• Solve this system for C1, C2...
• Can’t do it. Why? e3
3
e
e−3
C1
y0
=
−3
e
C2
v0 Independence and the Wronskian (Section 3.2)
y1 (t) = e2t+3 and y2 (t) = e2t−3 are two
• Example: Suppose
solutions to some equation. Can we solve ANY initial
y
condition . (0) = y0 , y (0) = v0 with these two solutions?
. y (t) = C1 e2t+3 + C2 e2t−3
y (0) = C1 e3 + C2 e−3 = y0 y (0) = 2C1 e3 + 2C2 e−3 = v0
• Solve this system for C1, C2...
• Can’t do it. Why?
e3 e−3
C1
y0
=
3
−3
ee
C2
v0
3
e e−3
det 3
=0
−3
ee Independence and the Wronskian (Section 3.2)
• For any two solutions to some linear ODE, to ensure that we have a
general solution, we need to check that
y1 (0) y2 (0)
det
= y1 (0)y2 (0) − y1 (0)y2 (0) = 0
y1 (0) y2 (0) Independence and the Wronskian (Section 3.2)
• For any two solutions to some linear ODE, to ensure that we have a
general solution, we need to check that
y1 (0) y2 (0)
det
= y1 (0)y2 (0) − y1 (0)y2 (0) = 0
y1 (0) y2 (0) Independence and the Wronskian (Section 3.2)
• For any two solutions to some linear ODE, to ensure that we have a
general solution, we need to check that
y1 (0) y2 (0)
det
= y1 (0)y2 (0) − y1 (0)y2 (0) = 0
y1 (0) y2 (0)
• For ICs other t...
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 Spring '13
 EricCytrynbaum
 Differential Equations, Equations, Complex Numbers

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