Section A.2 Linear Ordinary Diﬀerential Equations with Constant CoeﬃcientsA17961.To solvey±±-4y=0,y(0) = 0±(0) = 3, start with the general solutiony(x)=c1cosh 2x+c2sinh 2x.Theny(0) = 0⇒c1cosh 0 +c2sinh 0 = 0⇒c1= 0; soy(xc2sinh 2x.y±(0) = 3⇒2c2cosh 0 = 3⇒c2=32⇒y(x32sinh 2x.65.To solvey±±-5y±+6y=ex(0) = 0±(0) = 0, use the general solutionfrom Exercise 27 (modify it slightly):y=c1e2x+c2e3x+12ex.Theny(0) = 0⇒c1+c2=-12;y±(0) = 0⇒2c1+3c2=-12⇒c2=12;c1=-1⇒y(x-e2x+12e3x+12ex.69.Because of the initial conditions, it is more convenient to takey=c1cos[2(x-π2)] +c2sin[2(x-π2)]as a general solution of
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.