Answer to Question #125359 in Differential Equations for jse

Question

1. Find the general solution of the differential equation.
y′′′−6y′′−36y′+216y=0. Use C1, C2, C3, ... for the constants of integration.

y(t) = _______

2. Determine a suitable form for Y(t) if the method of undetermined coefficients is to be used.
Do not evaluate the constants.

y′′′−4y′ = te^−2t + 3cos(2t)

Use J,K,L,M as coefficients. Enclose arguments of functions in parentheses. For example, sin(2x). Do not simplify trigonometric functions of nt, where n is a positive integer.

Answer 1

Solution:

1.y′′′−6y′′−36y′+216y=0

k³-6k²-36k+216=0

k²(k-6)-36(k-6)=0

(k-6)(k^2-36)=0

(k-6)²(k+6)=0

k₁=k₂=6,k₃=-6

y(t)=C₁e^6t+C₂te^6t+C₃e^-6t

2.y′′′−4y′ = te^-2t + 3cos(2t)

k³-4k=0

k(k²-4)=0

k(k-2)(k+2)=0

k₁=-2, k₂=0, k₃=2

y₀(t)=C₁e^-2t+C₂+C₃e^2t

y_p1(t)=(Jt²+Kt)e^-2t

y_p2(t)=Lcos(2t)+Msin(2t)

y(t)=C₁e^-2t+C₂+C₃e^2t+(Jt²+Kt)e^-2t+Lcos(2t)+Msin(2t)

Answer:

1.y(t)=C₁e^6t+C₂te^6t+C₃e^-6t

2.y(t)=C₁e^-2t+C₂+C₃e^2t+(Jt²+Kt)e^-2t+Lcos(2t)+Msin(2t)