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Math 8 - W#3 42 f 6 First solve the differential equation...

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Unformatted text preview: W #3 42 f 6. First, solve the differential equation y” + cy’ — ay + by3 = A cos wt for the highest derivative of y present in the equation. )1” : ——-cy/ + ay —— by3 + A COS wt Next, set v = y’. Then 11' = y” 2 —cy’ + ay —— by3 + Acoswt ——cv + ay —— by3 + Acos cot. Thus, we now have the following system of first or- der equations. ' y’=v v’ = mcv +ay —by3 + Acoswt 12. If y” + 2y’ +17y = 0, then the characteristic equa— tion is A2+ZA+17=0. The roots of the characteristic equation are —1 :1: 4i , leading to the complex solutions z(t) = e(—1+4i)t and E(t) = e(—l~4i)t- However, by Euler’s identity, ~t 4ft z(t) : e e = e~’(cos4z + i sin 4!), and the, real and imaginary parts of 2 lead to a fun~ damental set of real solutions y1(t) = e” cos 4t and y2 (t) = e“’ sin 4:. Hence the general solution is y(t) 2 Ge" cos 4t + C2?" sin 4t. 20. If 4y” + 12y’ + 9y 2 0, then the characteristic equa— tion is 4k2+12l+9=(2)~+3)2=0~ Hence, the characteristic equation has a repeated root, A = —3/2. Therefore, y1(t) = 670/2)! and y2 (t) = Ira/2” form a fundamental set of real so— lutions. Hence, the general solution is y(t) = Clert3/2)‘+c2te*<3/2)’ = (Ci+C2z)e—<3/2>‘. ,3 6 Let y = e“ in 6y” + y’ — y = 0 to obtain 6’26“ +A‘elt __ eM = 0 ems)? + A _ 1) = 0. Because e“ .,+— 0, we arrive at the characteristic equa- tlon 6A2+A—1=0 (3i — 1)(2A + 1) = 0, and roots 1 = 1/3 and A = ——1/2. Because the - l 3 roots are distinct, the solutions y1(t) = e( / )‘ ‘and mm = e‘fl/Z)‘ form a fundamental set of solutions and the general solution is y(:) = Cid/3 + Cir/2. If 10y” ~ 12’ — 3y = 0, then the characteristic equa~ tion is 26. 10A2~A~3=(2A+1)(5)i——3)=0, with roots A = —~1/2 and A = 3/5. This leads to the general solution y(t) = Cle—t/2 + C2e3’/5. Using the initial condition y(0) = 1 provides 1 = C1 + C2. Differentiating the general solution, 1 3 y/(t) = ——C1e“‘/2 + ~C2e3'/5, 2 5 ‘f‘len using the initial condition y’(0) = 0 leads to 1 3 o = ——C —C . 2 -‘ + 5 2 7‘hese equations yield C1 = 6/11 and C2 = 5/ 1], giving the particular solution 6 I ._ —/2 116 + 5 ——-e 3t/5 11 ‘ y(t) = ...
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