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use the annihilator method to determinethe form of a particular solution for the given equation.u''-5u'+6u=cos2x+1

Short Answer

Expert verified

up(x)=c3+c4sin2x+c5cos2x

Step by step solution

01

Solve the homogeneous of the given equation

The homogeneous of the given equation is

D2-5D+6[u]=0⇒(D-2)(D-3)[u]=0

The solution of the homogeneous is

uh(x)=c1e2x+c2e3x (1)

Letg(x)=cos2x

Then

(D2+22)[g]=0⇒(D2+4)[g]=0

Lethx=1

Then

D[h]=0

Hence

D(D2+4)[g+h]=0

Then, every solution to the given nonhomogeneous equation also satisfies

DD2+4(D-2)(D-3)[u]=0

Then

u(x)=c1e2x+c2e3x+c3+c4sin2x+c5cos2x(2)

is the general solution to this homogeneous equation

We knowu(x)=uh+up

Comparing (1) & (2)

up(x)=c3+c4sin2x+c5cos2x

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Most popular questions from this chapter

find a differential operator that annihilates the given function.

x2-ex

Use the annihilator method to show that ifa0=0anda1≠0in (4) andhas the form f(x)given in (17), then equation (4) has a particular solution of the formyp(x)=x{Bmxm+Bm-1xm-1+⋯+B1x+B0}

Determine whether the given functions are linearly dependent or linearly independent on the specified interval. Justify your decisions.

{cos2x, cos2x, sin2x}on(-∞, ∞)

As an alternative proof that the functions er1x,er2x,er3x,...,ernxare linearly independent on (∞,-∞) when r1,r2,...rn are distinct, assume C1er1x+C2er2x+C3er3x+...+Cnernxholds for all x in (∞,-∞) and proceed as follows:

(a) Because the ri’s are distinct we can (if necessary)relabel them so that r1>r2>r3>...>rn.Divide equation (33) by to obtain C1+C2er2xer2x+C3er3xer3x+...+Cnernxernx=0Now let x→∞ on the left-hand side to obtainC1 = 0.(b) Since C1 = 0, equation (33) becomes

C2er2x+C3er3x+...+Cnernx= 0for all x in(∞,-∞). Divide this equation byer2x

and let x→∞ to conclude that C2 = 0.

(c) Continuing in the manner of (b), argue that all thecoefficients, C1, C2, . . . ,Cn are zero and hence er1x,er2x,er3x,...,ernxare linearly independent on(∞,-∞).

Deflection of a Beam Under Axial Force. A uniform beam under a load and subject to a constant axial force is governed by the differential equation

y(4)(x)-k2y''(x)=q(x),0<x<L,

where is the deflection of the beam, L is the length of the beam, k2is proportional to the axial force, and q(x) is proportional to the load (see Figure 6.2).

(a) Show that a general solution can be written in the form

y(x)=C1+C2x+C3ekx+C4e-kx+1k2∫q(x)xdx-xk2∫q(x)dx+ekx2k3∫q(x)e-kxdx-e-kx2k3∫q(x)ekxdx

(b) Show that the general solution in part (a) can be rewritten in the form

y(x)=c1+c2x+c3ekx+c4e-kx+∫0xq(s)G(s,x)ds,

where

G(s,x):=s-xk2-sinh[k(s-x)]k3.

(c) Let q(x)=1 First compute the general solution using the formula in part (a) and then using the formula in part (b). Compare these two general solutions with the general solution

y(x)=B1+B2x+B3ekx+B4e-kx-12k2x2,

which one would obtain using the method of undetermined coefficients.

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