ES.1803 Topic 19 Notes Jeremy Orloff. 19 Variation of parameters; exponential inputs; Euler s method

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1 ES83 Topic 9 Notes Jeremy Orloff 9 Variation of parameters; eponential inputs; Euler s method 9 Goals Be able to derive and apply the eponential response formula for constant coefficient linear systems with eponential input 2 Be able to solve linear constant coefficient systems with sinusoidal input using comple replacement and the ERF 3 Be able to use the variation of paramters formula to solve a (nonconstant) coefficient linear inhomogeneous system 4 Be able to use Euler s method to approimate the solution to a system of first order equations 92 Introduction We now turn our attention to inhomogeneous linear systems of the form A(t) + F(t) (I) Here is a column vector of (unknown) functions, A(t) is a square matri and the input F(t) is a column vector The associated homogeneous equation is A(t) (H) In this topic we will first consider the case where the coefficient matri A is constant and the input F is eponential Perhaps unsurprisingly, this will lead to the eponential response formula for a solution As with ordinary differential equations, this will also allow us to find the response to sinusoidal input Net we will look at linear equations with arbitrary input This will lead to the variation of parameters formula for the solution This is a beautiful formula, it can be painful or difficult to apply We will use it as a last resort to look for solutions to equations with nonconstant coefficients or unusual input 93 Eponential response formula (ERF) Eponential response formula For a constant matri A and a constant vector k the DE A + e at k has a particular solution: p e at (ai A) k

2 9 VARIATION OF PARAMETERS; EXPONENTIAL INPUTS; EULER S METHOD2 This formula is valid as long as ai A is invertible, ie as long as a is not an eigenvalue of A Proof Not surprisingly we discover this formula by the method of optimism We try a solution of the form p (t) e at v, where v is a constant vector Plug the guess into the DE and solve for v: p ae at v e at Av + e at k (ai A)v k v (ai A) k Thus, we have found a particular solution p (t) e at v e at (ai A) k QED 6 5 e 2t Eample 9 Find the general solution to y + 2 y 3e 2t answer: For ease of notation we rewrite the equation as A+e 2t The eponential 3 response formula gives us a particular solution p (t) e 2t (2I A) 3 e 2t e2t e2t We know from previous topics that the general homogeneous equation is ] ] h (t) c e t [ + c 2 e 7t [ 5 By superposition the general solution to the system is (t) p (t) + h (t) 6 5 3e 2t Eample 92 Solve y + 2 y 5e 3t 3 answer: Write the input as e 2t + e 3t Now you can solve the equation for each 5 input term and then use superposition There are more eamples in the net section 94 Eponential response formula eamples Eample 93 Find the general solution to 3 y + e 2t y 4 y e 2t

3 9 VARIATION OF PARAMETERS; EXPONENTIAL INPUTS; EULER S METHOD3 answer: In matri form the equation is response formula tells us a particular solution is p e 2t (2I A) [ ] 3 + e 4 2t The eponential e 2t e 2t 3 e 2t We ll let you verify the calculation of the inverse Likewise we ll let you find the homogeneous solution needed for the general solution Eample 94 Find a particular solution to 3 y + 3 y 4 y + 2 answer: Note that we could phrase this using the eponential response formula, where the eponent is a Instead we ll just say that we re guessing a constant solution and solve for its eact value Try v Substitution into the DE gives Av So, v A That is 6 p (t) 6 Again, we ll let you verify the calculation of the inverse 2 cos(t) Eample 95 Find a particular solution to + 2 y answer: To use the eponential response formula we first need to use comple replacement The compleified equation is z 2 z + e it, where Re(z) 2 Now we set up the eponential response formula (ii A) + i i So, z p e it (ii A) y [ + i 2 2i 4 eit 2 + i + i 2 2i i ] 2i + 4 eit i 2 To find the real part of z p we work in polar coordinates First we write the various comple numbers in polar form: 2i e iφ, where φ Arg(2i + 4) tan (5) in the first quadrant Likewise i 2e iφ2;, where φ 2 π/4 [ ] So z p eit 2e 2 iφ 2 [ ] 2e 5e iφ 2 2 i(t+φ 2 φ ) 5 2e i(t φ ) Taking the real part: p Re(z p ) [ ] 2 cos(t + φ2 φ ) 2 cos(t φ )

4 9 VARIATION OF PARAMETERS; EXPONENTIAL INPUTS; EULER S METHOD4 Here is the same calculation in rectangular coordinates I think the arithmetic is more error prone and the answer is harder to interpret i 4 2i i 3i 2i i So, z p (cos(t) + i sin(t)) Thus, 3i [ 4 + 2i p (t) Re(z p (t)) cos(t) + 3 sin(t) + i(sin(t) 3 cos(t)) 4 cos(t) 2 sin(t) + i( 4 sin(t) + 2 cos(t)) cos(t) + 3 sin(t) 4 cos(t) 2 sin(t) ] 95 Variation of parameters For the general, not necessarily constant coefficient, linear inhomogeneous system (I) we cannot use constant coefficient techniques like the ERF For those cases where the ERF doesn t apply we can try to use the variation of parameters formula Since it involves integration, matri inverses and matri multiplication it is our last choice when trying to solve an equation Nonetheless, sometimes it s the only method available In addition, the derivation of the formula is really very pretty Suppose we have a fundamental matri Φ(t) for the homogeneous linear equation A(t) (H) Remember this means that Φ has columns which are independent solutions to (H) 96 Variation of parameters formula Now suppose we want to solve A(t) + F(t) Theorem The general solution to equation (I) is given by the variation of parameters formula ( ) (t) Φ(t) Φ(t) F(t) dt + C Proof (Remember this) We will use a form of the method of optimism to derive this formula We know the general homogeneous solution is (t) Φ(t) c for a constant vector c The vector c is called a parameter Variation of parameters is an old-fashioned way of saying let s optimistically make it a (dependent) variable u(t) So, we try a solution of the form (t) Φ(t) u(t) The function u(t) is unknown, we substitute our guess into (I) and see what u(t) needs to be: Φ u + Φ u AΦ u + F So, (don t forget Φ AΦ) AΦ u + Φ u AΦ u + F Φ u F (I)

5 9 VARIATION OF PARAMETERS; EXPONENTIAL INPUTS; EULER S METHOD5 This last equation is easy to solve u Φ F u(t) Φ (t) F(t) dt + C We take this formula for u(t) and use it in our trial solution ( ) (t) Φ(t) u(t) Φ(t) Φ(t) F(t) dt + C QED Remark Note that the variation of parameters formula assumes you know the general homogeneous solution It gives no help in finding this solution (For eamples we can use constant coefficient systems where we know how to find the homogeneous solution using eigenvalues and eigenvectors) Eample 96 Use the variation of parameters formula to solve 6 5 e t + 2 e 5t Note We retiterate that using the ERF is the preferred method of solving this equation We use the variation of parameters formula here for practice 6 5 answer: Let s introduce some notation to save typing: A, F 2 t [ e t 5e We know the fundamental matri from an earlier eample: Φ(t) 7t ] e t e 7t So, [ e Φ 7t 5e (t) 7t ] e 8t 6 e t e t Calculating with the variation of parameters we get Φ(t) Φ (t) F(t) dt [ e 8t e 7t 5e Φ(t) 7t ] e t 6 e t e t e 5t dt 5e 4t Φ(t) 6 e 6t + e 2t dt [ 6 Φ(t) t 5 ] 4 e4t + c 6 e 6t 2 e 2t + c 2 [ te t 5 4 e5t 5 6 et 5 2 e5t + c e t + 5c 2 e 7t ] 6 te t e5t 6 et 2 e5t c e t + c 2 e 7t [ ] te t + e 5t 5/4 + e t 5/6 + c 6 3/4 /6 e t + c 2 e 7t 5 Notice the homogeneous solution appearing with the constants of integration

6 9 VARIATION OF PARAMETERS; EXPONENTIAL INPUTS; EULER S METHOD6 96 Definite integral version of variation of parameters The equation (I) with initial condition (t ) b has definite integral solution 97 Euler s method ( t ) (t) Φ(t) Φ (u) F(u) du + C where C Φ (t ) b t Euler s method works without change for systems Consider a first order system: F(, t), (t ) Using stepsize h we have the same algorithm as for ordinary DE s m F( n, t n ) n+ n + hm, t n+ t n + h Again, just like for ordinary DE s, there are other, better, algorithms for choosing m or varying h y Eample 97 Consider y t, () Let and use h 5 to estimate (2) y n t n n m F( n, t n )