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chapter2.page15

# chapter2.page15 - t 2 x 3 3 t-4 x we have ∂f t,x ∂t = 2...

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4. INITIAL VALUE PROBLEM AND EXISTENCE THEOREM 15 with a ( t ) = 2 t and b ( t ) = 4 t 3 . So R a ( t ) dt = R 2 t dt = t 2 , and x ( t ) = e - t 2 ˆ Z 4 t 3 e t 2 dt + C ! . Using method of substitution or Mathcad , we find x ( t ) = e - t 2 2 t 2 e t 2 - 2 e t 2 + C Let t = 0 we have x (0) = C - 2 . From initial condition x (0) = 1 we see that C = 3 . So the solution is x ( t ) = e - t 2 2 t 2 e t 2 - 2 e t 2 + 3 a However, for general first order ODE , the problem is not so nice, the following theorem is a little bit hard to state and to understand. Theorem 4.2 . Let f ( t, x ) be defined on interval I. If the partial derivative of f ( t, x ) with respect to y (denoted as ∂f ( t,x ) ∂y , reads ”par- tial f(t,x) over partial y”) and f ( t, x ) are continuous at an open disk centered at ( t 0 , x 0 ) , then the initial value problem x 0 = f ( t, x ) x ( t 0 ) = x 0 has an unique solution on some open interval containing the point t 0 . Remark 4.1 . We will not discuss this existence and uniqueness theorem for general ODE . But we would like to make some comments. - To find ∂f ( t,x ) ∂t , you just need to treat x variable as a constant, for example, let
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Unformatted text preview: t 2 x 3 + 3 t-4 x, we have ∂f ( t,x ) ∂t = 2 tx 3 +3 , here 4 x is a constant with respect to t , so the derivative is 0, and x 3 is also a constant, so t 2 x 3 when taking derivative against t we have 2 tx 3 . Therefore ﬁnding partial derivative is as easy as to ﬁnd ordinary derivative.-This theorem only guarantees a solution deﬁned for t in an open interval, which might be a bounded interval. For example x = x 2 , we see that x ( t ) =-1 t + c is the general solution, if c =-2 then the solution is only deﬁned on interval (0, 2) as x ( t ) =-1 t-2 will become undeﬁned at t = 2 .-Some time we could have more than one solution that satisﬁes the same initial condition, for example, for x = x 2 3 and x (0) = we have two diﬀerent solution, one is x ( t ) = 0 another one...
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