Lectur16 - Lecture XVI More Complicated Solutions to...

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Lecture XVI More Complicated Solutions to Differential Equations I. Nonlinear Differential Equations A. In fairly general terms, we can assert that a first-order linear differential equation has a unique solution. This assertion follows our solution of the & () xp txg t += problem. B. However, solutions for nonlinear differential equations where & (, ) xf t x = are somewhat more difficult and lack a general solution framework. For example, & xx x == 1 3 00 raises some additional difficulties. C. Two procedures for solving nonlinear differential equations: 1. Separable Differential Equations a. Given the general expression of a differential equation: dx dt ftx = an alternative way to express this relationship is Mtx Ntx 0 . One subgroup of this class of differential equations is referred to as separable differential equations where Mt Nx ( ) 0 so that each term only involves t or x respectively. Solving this system then involves integrating each section separately: Mt Nx Mtdt Nxd x M t dt N x dx =- = = ∫∫ b. Example 1: tt x x = ++ - 34 2 21 01 2 ,() Separating the function
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(29 21 34 2 3 4 2 2 1 2 22 2 2 2 23 2 2 ()( ) () ( ) xd xt td t zd z s sd s c xx t t t c ttt c -=+ + -= + ++ - =+ + + + ∫∫ Given that x(0) = -1, 121 3 --=⇒= cc Thus, the implicit form of the solution is 2 2 3 + + + c. Example 2: dx dt t x = 2 Separating the equation: xdx t dt z dz s ds c x t c c = 2 2 2 3 1 2 3 2. Exact Differential Equations a. A second useful technique starts from a general equation ψ (, ) tx c = taking the total differential of this equation with dc = 0 we obtain: ψψ tx t x dt t x dx += 0 0 or Obviously, we see that since ψ s t z x t x dt t x dx d t x sxd s tzd z = = or Spotting this symmetry between the terms allows for the solution of these problems.
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Lectur16 - Lecture XVI More Complicated Solutions to...

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