Existence and Uniqueness Theorem
- y′′+p(t)y′+q(t)y=g(t), y(t0)y0,y′(t0)=y0′
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where p,q and g are continuous on an open interval and it contains the point t0,
- that means the problem has exactly one solution y=ϕ(t) and the solution exists in the same interval
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What does this mean?
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- the IVP must have a solution (a solution exists)
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- the IVP has only one solution (the solution is unique)
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- the solution ϕ is defined throughout the interval I where everything is continuous and 2x differentiable
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Example
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find the longest interval in which the solution of the IVP is certain to exist
- (t2−3t)y′′+ty′−(t+3)y=0, y(1)=2 y′(1)=1
- if the equation is written of the form as the definition
- p(t)=t−31,q(t)=−t(t−3)t+3 and g(t)=0
- we can see there are discontinuities at t=0 and 3, therefore the longest open interval the solution can exist in is 0<t<3
The Principle of Superposition
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when you have 2 solutions, a linear combination of the 2 will also be a solution