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Question Number 184306 by Mastermind last updated on 05/Jan/23

Consider the boundary value   problem y^(′′) −2y′+2y=0,       y(a)=c  ,y(b)=d.  1) If this problem has a unique  solution, how are a and b related?  2) If this problem has no solution,  how are a,b,c and d related?      Help!

$$\mathrm{Consider}\:\mathrm{the}\:\mathrm{boundary}\:\mathrm{value}\: \\ $$$$\mathrm{problem}\:\mathrm{y}^{''} −\mathrm{2y}'+\mathrm{2y}=\mathrm{0},\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{a}\right)=\mathrm{c} \\ $$$$,\mathrm{y}\left(\mathrm{b}\right)=\mathrm{d}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{If}\:\mathrm{this}\:\mathrm{problem}\:\mathrm{has}\:\mathrm{a}\:\mathrm{unique} \\ $$$$\mathrm{solution},\:\mathrm{how}\:\mathrm{are}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{related}? \\ $$$$\left.\mathrm{2}\right)\:\mathrm{If}\:\mathrm{this}\:\mathrm{problem}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solution}, \\ $$$$\mathrm{how}\:\mathrm{are}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{and}\:\mathrm{d}\:\mathrm{related}? \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Answered by FelipeLz last updated on 05/Jan/23

y′′−2y′+2y = 0  y = e^r  → r^2 e^r −2re^r +2e^r  = 0  r^2 −2r+2 = 0 → r = 1±i  y(x) = k_1 e^x cos(x)+k_2 e^x sin(x)     1. y(a) = c → k_1 e^a cos(a)+k_2 e^a sin(a) = c       y(b) = d → k_1 e^b cos(b)+k_2 e^b sin(b) = d        determinant (((e^a cos(a)),(e^a sin(a))),((e^b cos(b)),(e^b sin(b))))≠ 0       e^(a+b) cos(a)sin(b)−e^(a+b) sin(a)cos(b) ≠ 0       e^(a+b) sin(b−a) ≠ 0       sin(b−a) ≠ 0       b ≠ a+nπ  ∀n ∈ Z      2.  determinant (((e^a cos(a)),(e^a sin(a))),((e^b cos(b)),(e^b sin(b))))= 0       e^(a+b) cos(a)sin(b)−e^(a+b) sin(a)cos(b) = 0       e^(a+b) sin(b−a) = 0       sin(b−a) = 0       b = a+nπ  ∀n ∈ Z        determinant ((c,(e^a sin(a))),(d,(e^b sin(b))))≠ 0       ce^b sin(b)−de^a sin(a) ≠ 0       ce^(a+nπ) sin(a+nπ)−de^a sin(a) ≠ 0       ce^(a+nπ) sin(a)cos(nπ)−de^a sin(a) ≠ 0       e^a sin(a)[ce^(nπ) cos(nπ)−d] ≠ 0       d ≠ ce^(nπ) cos(nπ)        { ((b−a = nπ  ∀n ∈ Z)),((d ≠ ce^(b−a) cos(b−a))) :}

$${y}''−\mathrm{2}{y}'+\mathrm{2}{y}\:=\:\mathrm{0} \\ $$$${y}\:=\:{e}^{{r}} \:\rightarrow\:{r}^{\mathrm{2}} {e}^{{r}} −\mathrm{2}{re}^{{r}} +\mathrm{2}{e}^{{r}} \:=\:\mathrm{0} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{r}+\mathrm{2}\:=\:\mathrm{0}\:\rightarrow\:{r}\:=\:\mathrm{1}\pm{i} \\ $$$${y}\left({x}\right)\:=\:{k}_{\mathrm{1}} {e}^{{x}} \mathrm{cos}\left({x}\right)+{k}_{\mathrm{2}} {e}^{{x}} \mathrm{sin}\left({x}\right) \\ $$$$\: \\ $$$$\mathrm{1}.\:{y}\left({a}\right)\:=\:{c}\:\rightarrow\:{k}_{\mathrm{1}} {e}^{{a}} \mathrm{cos}\left({a}\right)+{k}_{\mathrm{2}} {e}^{{a}} \mathrm{sin}\left({a}\right)\:=\:{c} \\ $$$$\:\:\:\:\:{y}\left({b}\right)\:=\:{d}\:\rightarrow\:{k}_{\mathrm{1}} {e}^{{b}} \mathrm{cos}\left({b}\right)+{k}_{\mathrm{2}} {e}^{{b}} \mathrm{sin}\left({b}\right)\:=\:{d} \\ $$$$\:\:\:\:\:\begin{vmatrix}{{e}^{{a}} \mathrm{cos}\left({a}\right)}&{{e}^{{a}} \mathrm{sin}\left({a}\right)}\\{{e}^{{b}} \mathrm{cos}\left({b}\right)}&{{e}^{{b}} \mathrm{sin}\left({b}\right)}\end{vmatrix}\neq\:\mathrm{0} \\ $$$$\:\:\:\:\:{e}^{{a}+{b}} \mathrm{cos}\left({a}\right)\mathrm{sin}\left({b}\right)−{e}^{{a}+{b}} \mathrm{sin}\left({a}\right)\mathrm{cos}\left({b}\right)\:\neq\:\mathrm{0} \\ $$$$\:\:\:\:\:{e}^{{a}+{b}} \mathrm{sin}\left({b}−{a}\right)\:\neq\:\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{sin}\left({b}−{a}\right)\:\neq\:\mathrm{0} \\ $$$$\:\:\:\:\:{b}\:\neq\:{a}+{n}\pi\:\:\forall{n}\:\in\:\mathbb{Z}\: \\ $$$$\: \\ $$$$\mathrm{2}.\:\begin{vmatrix}{{e}^{{a}} \mathrm{cos}\left({a}\right)}&{{e}^{{a}} \mathrm{sin}\left({a}\right)}\\{{e}^{{b}} \mathrm{cos}\left({b}\right)}&{{e}^{{b}} \mathrm{sin}\left({b}\right)}\end{vmatrix}=\:\mathrm{0} \\ $$$$\:\:\:\:\:{e}^{{a}+{b}} \mathrm{cos}\left({a}\right)\mathrm{sin}\left({b}\right)−{e}^{{a}+{b}} \mathrm{sin}\left({a}\right)\mathrm{cos}\left({b}\right)\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:{e}^{{a}+{b}} \mathrm{sin}\left({b}−{a}\right)\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{sin}\left({b}−{a}\right)\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:{b}\:=\:{a}+{n}\pi\:\:\forall{n}\:\in\:\mathbb{Z} \\ $$$$\:\:\:\:\:\begin{vmatrix}{{c}}&{{e}^{{a}} \mathrm{sin}\left({a}\right)}\\{{d}}&{{e}^{{b}} \mathrm{sin}\left({b}\right)}\end{vmatrix}\neq\:\mathrm{0} \\ $$$$\:\:\:\:\:{ce}^{{b}} \mathrm{sin}\left({b}\right)−{de}^{{a}} \mathrm{sin}\left({a}\right)\:\neq\:\mathrm{0} \\ $$$$\:\:\:\:\:{ce}^{{a}+{n}\pi} \mathrm{sin}\left({a}+{n}\pi\right)−{de}^{{a}} \mathrm{sin}\left({a}\right)\:\neq\:\mathrm{0} \\ $$$$\:\:\:\:\:{ce}^{{a}+{n}\pi} \mathrm{sin}\left({a}\right)\mathrm{cos}\left({n}\pi\right)−{de}^{{a}} \mathrm{sin}\left({a}\right)\:\neq\:\mathrm{0} \\ $$$$\:\:\:\:\:{e}^{{a}} \mathrm{sin}\left({a}\right)\left[{ce}^{{n}\pi} \mathrm{cos}\left({n}\pi\right)−{d}\right]\:\neq\:\mathrm{0} \\ $$$$\:\:\:\:\:{d}\:\neq\:{ce}^{{n}\pi} \mathrm{cos}\left({n}\pi\right) \\ $$$$\:\:\:\:\:\begin{cases}{{b}−{a}\:=\:{n}\pi\:\:\forall{n}\:\in\:\mathbb{Z}}\\{{d}\:\neq\:{ce}^{{b}−{a}} \mathrm{cos}\left({b}−{a}\right)}\end{cases} \\ $$$$\:\:\:\:\: \\ $$

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