1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a nonconstant can be solved with the same idea as above.


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1 1 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt Coefficient Equtions Remrk: Summry of the constnt coefficient cse: y = y + b We wrote the eqution y = y + b s follows y = y + b " The criticl step ws the following: since b/ is constnt, then (b/) = 0, hence y + b " = y + b " At this point the eqution ws simple to solve, (y + b # ) ##y (y + b ) = ln b " # ##y b " + # = ln + # = c 0 + t We now computed the exponentil on both sides, to get # #y + b # = e c0+t = e c 0 e t y + b = (±ec 0 ) e t, nd clling c = ±e c 0 we got the formul y(t) = c e t b, Remrk: The cse b = 0 nd nonconstnt cn be solved with the sme ide s bove y y = (t) ln( y ) = (t) ln( y(t) ) = A(t) + c 0, where A = $ dt, is primitive or ntiderivtive of Therefore, y(t) = ±e A(t)+c 0 = ±e A(t) e c 0 y(t) = c e A(t), c = ±e c 0 Exmple 121: The solutions of y = 2t y re y(t) = c e t2, where c R Remrk: The cse b/ nonconstnt cnnot be solved with this ide
2 2 122 Solving Vrible Coefficient Equtions Theorem 121 (Vrible Coefficients) If the functions, b re continuous, then y = (t) y + b(t), (121) hs infinitely mny solutions given by y(t) = c e A(t) + e A(t) e A(t) b(t) dt, (122) where A(t) = " (t) dt nd c R Remrks: () The expression in Eq (122) is clled the generl solution (b) The function µ(t) = e A(t) is the integrting fctor Exmple 122: Show tht for constnt coefficient equtions the solution formul given in Eq (122) reduces to Eq (114) Solution: In the prticulr cse of constnt coefficient equtions, primitive, or ntiderivtive, for the constnt function is A(t) = t, so y(t) = c e t + e t e t b dt Since b is constnt, the integrl in the second term bove cn be computed explicitly, e t b e t dt = e t# b e t$ = b Therefore, in the cse of, b constnts we obtin y(t) = c e t b given in Eq (114)
3 3 Proof of Theorem 121: Rewrite: y y = b, Multiply nd then multiply by µ, n integrting fctor, µ y µ y = µ b (123) The criticl step: choose µ such tht µ = µ (124) Why? Becuse for ny µ solution of Eq (124) we hve µ y + µ y = µ b, nd the lefthnd side is totl derivtive of product of two functions, µ y " = µ b (125) Tht s why Now solve for µ, µ = µ µ µ = ln( µ ) = ln( µ ) = A + c 0, where A = # dt, primitive or ntiderivtive of, nd c 0 is n rbitrry constnt Then, µ = ±e c 0 e A µ = c 1 e A, c 1 = ±e c 0 We choose c 0 = 0, hence c 1 = 1 The integrting fctor is then µ(t) = e A(t) This function is n integrting fctor, becuse if we strt gin t Eq (123), we get e A y e A y = e A b e A y + e A" y = e A b,
4 4 where we used the min property of the integrting fctor, e A = e A" Now the product rule for derivtives implies tht the lefthnd side bove is totl derivtive, e A y " = e A b Integrting on both sides we get e A y " # = e A b dt + c e A y " # e A b dt = c The function ψ(t, y) = e A y " $ e A b dt is clled potentil function of the differentil eqution The solution of the differentil eqution is # y(t) = c e A(t) + e A(t) e A(t) b(t) dt This estblishes the Theorem
5 5 Exmple 124: Find ll solutions of ty = 2y + 4t 2, with t > 0 Solution: Rewrite the eqution s y = 2 t y + 4t (t) = 2, b(t) = 4t (126) t Rewrite gin, y + 2 t y = 4t Multiply by function µ, µ y + 2 t µ y = µ 4t Choose µ solution of 2 t µ = µ ln( µ ) = 2 t ln( µ ) = 2 ln(t) = ln(t 2 ) µ(t) = ±t 2 We choose µ = t 2 Multiply the differentil eqution by this µ, t 2 y + 2t y = 4t t 2 (t 2 y) = 4t 3 If we write the righthnd side lso s derivtive, t 2 y " = t 4 " t 2 y t 4" = 0 So potentil function is ψ(t, y(t)) = t 2 y(t) t 4 Integrting on both sides we obtin t 2 y t 4 = c t 2 y = c + t 4 y(t) = c t 2 + t2 Exmple 125: Find the solution to the initil vlue problem ty + 2y = 4t 2, t > 0, y(1) = 2 Solution: The generl solution is y(t) = c t 2 + t2 The initil condition implies tht 2 = y(1) = c + 1 c = 1 y(t) = 1 t 2 + t2
6 6 123 The Bernoulli Eqution Definition 124 The Bernoulli eqution is y = p(t) y + q(t) y n, n R, (127) Remrks: () For n = 0, 1 the eqution is nonliner (b) If n = 2 we get the logistic eqution, (we ll study it in lter chpter), y = ry 1 y " K (c) This is not the Bernoulli eqution from fluid dynmics (d) The Bernoulli eqution is nonliner eqution tht cn be trnsformed into liner eqution Theorem 125 (Bernoulli) The function y is solution of the Bernoulli eqution y = p(t) y + q(t) y n, n = 1, iff the function v = 1/y (n 1) is solution of the liner differentil eqution v = (n 1)p(t) v (n 1)q(t) Proof of Theorem 125: Divide the Bernoulli eqution by y n, y y n = p(t) + q(t) yn 1 Introduce the new unknown v = y (n 1) nd compute its derivtive, v = # y (n 1)$ = (n 1)y n y v (t) (n 1) = y (t) y n (t) If we substitute v nd this lst eqution into the Bernoulli eqution we get v (n 1) = p(t) v + q(t) v = (n 1)p(t) v (n 1)q(t) This estblishes the Theorem
7 7 Exmple 127: Find every nonzero solution of the differentil eqution y = y + 2y 5 Solution: This is Bernoulli eqution for n = 5 Divide the eqution by the nonliner fctor y 5, y y 5 = 1 y Introduce the function v = 1/y 4 nd its derivtive v = 4(y /y 5 ), into the differentil eqution bove, v 4 = v + 2 v = 4 v 8 v + 4 v = 8 The lst eqution is liner differentil eqution for the function v This eqution cn be solved using the integrting fctor method Multiply the eqution by µ(t) = e 4t, then e 4t v " = 8 e 4t e 4t v = 8 4 e4t + c We obtin tht v = c e 4t 2 Since v = 1/y 4, 1 y 4 = c 1 e 4t 2 y(t) = ± c e 4t 2 " 1/4
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