Please solve d and e only solution of a,b,c is provided Case(A) We have to find out the particularfunction of the form⇒ xp = A cos(45t) + B sin(45t)dNow xp& xp=dt= -45A sin(45t) + 45B cos(45t)d[-45A sin(45t) + 45B cos(45t)]dt= -2025A cos(45t) - 2025B sin(45t)=[A cos(45t) + B sin(45t)]Since xp will satisfy eqn(3) so we get,→ [-2025A cos(45t) - 2025B sin(45t)]+3025[A cos(45t) + B sin(45t)] = 500 cos(45t)⇒ 1000A cos(45t) + 1000B sin(45t) = 500 cos(45t)On comparison, we get⇒ 1000A = 500 & 1000B = 01⇒ A=2So required particular solution is given by1⇒ xp = cos(45t)2& B = 0Case(B) General solution of eqn(3) is given by⇒x=Xc + Xp1⇒ x = C₁ Cos(55t) + C₂ sin(55t) += cos(45t)2Case(C) We have x(0) = 0 so from eqn(4),1⇒0=C₁ +0+ ⇒C1=-2Then eqn(4) reduces into12Now x' =⇒x'=1⇒X=-- cos(55t) + C₂ sin(55t) + -cos(45t)2ddt552121cos(55t) + C2 sin(55t) + -cos(45t)245-sin(55t) +55c2 cos(55t) -—=sin(45t)2We have given that, x'(0) = 0 so from eqn(6)⇒ 0 = 0+55c₂-0 C₂ = 0Therefore required solution is given by,11⇒X=-- cos(55t) + cos(45t)22(4)(5)(6) 3025,6. Beats. Consider a free undamped oscillator with mass m = 1 and stiffness k =which satisfies the differential equationx"(t) + 3025x(t) = 0.The natural frequency is wo√k/m = 55 and the general solution is e(t) = C₁ cos(55t) +C₂ sin(55t). Now suppose we subject this oscillator to a periodic external force with amplitude500 and frequency 45:x"(t) + 3025x(t)500 cos(45t).(a) Find a particular solution of the form xp(t) = A cos(45t) + B sin(45t).(b) Find the general solution r(t) = xc(t) + xp(t).(c) Find the unique solution r(t) with initial conditions (0) = 0 and x'(0) = 0.(d) Express your solution in the form x(t) = C sin(at) sin(ßt). [Hint: Use the trig identities-cos(a - b) = cos a cos 3 + sin a sin 3,cos(a + B) = cos a cos - sin a sin 3,cos(a + 3) = 2 sin a sin 6.]cos(a - b)(e) Plot your solution r(t) for t between 0 and 3π/5. [Use a computer.]

Question

Please solve d and e only solution of a,b,c is provided Case(A) We have to find out the particularfunction of the form⇒ xp = A cos(45t) + B sin(45t)dNow xp& xp=dt= -45A sin(45t) + 45B cos(45t)d[-45A sin(45t) + 45B cos(45t)]dt= -2025A cos(45t) - 2025B sin(45t)=[A cos(45t) + B sin(45t)]Since xp will satisfy eqn(3) so we get,→ [-2025A cos(45t) - 2025B sin(45t)]+3025[A cos(45t) + B sin(45t)] = 500 cos(45t)⇒ 1000A cos(45t) + 1000B sin(45t) = 500 cos(45t)On comparison, we get⇒ 1000A = 500 & 1000B = 01⇒ A=2So required particular solution is given by1⇒ xp = cos(45t)2& B = 0Case(B) General solution of eqn(3) is given by⇒x=Xc + Xp1⇒ x = C₁ Cos(55t) + C₂ sin(55t) += cos(45t)2Case(C) We have x(0) = 0 so from eqn(4),1⇒0=C₁ +0+ ⇒C1=-2Then eqn(4) reduces into12Now x' =⇒x'=1⇒X=-- cos(55t) + C₂ sin(55t) + -cos(45t)2ddt552121cos(55t) + C2 sin(55t) + -cos(45t)245-sin(55t) +55c2 cos(55t) -—=sin(45t)2We have given that, x'(0) = 0 so from eqn(6)⇒ 0 = 0+55c₂-0 C₂ = 0Therefore required solution is given by,11⇒X=-- cos(55t) + cos(45t)22(4)(5)(6) 3025,6. Beats. Consider a free undamped oscillator with mass m = 1 and stiffness k =which satisfies the differential equationx&#34;(t) + 3025x(t) = 0.The natural frequency is wo√k/m = 55 and the general solution is e(t) = C₁ cos(55t) +C₂ sin(55t). Now suppose we subject this oscillator to a periodic external force with amplitude500 and frequency 45:x&#34;(t) + 3025x(t)500 cos(45t).(a) Find a particular solution of the form xp(t) = A cos(45t) + B sin(45t).(b) Find the general solution r(t) = xc(t) + xp(t).(c) Find the unique solution r(t) with initial conditions (0) = 0 and x'(0) = 0.(d) Express your solution in the form x(t) = C sin(at) sin(ßt). [Hint: Use the trig identities-cos(a - b) = cos a cos 3 + sin a sin 3,cos(a + B) = cos a cos - sin a sin 3,cos(a + 3) = 2 sin a sin 6.]cos(a - b)(e) Plot your solution r(t) for t between 0 and 3π/5. [Use a computer.]

Accepted Answer

(d)
Since, the general solution is of the form x = -12cos55t + 12cos45t
It can be simplified as:
x =…