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Chapter 6: Q4E (page 332)

Find a general solution for the differential equation with x as the independent variable:y'''鈭�3y''鈭�y'+3y=0

Short Answer

Expert verified

The general solution for the differential equation with x as the independent variableis.y(x)=c1e鈭�x+c2e鈭�5x+c3e4x

Step by step solution

01

Auxiliary equation:

The auxiliary equation isr3+2r2鈭�19r鈭�20=0

02

Inspecting the sum further:

By inspection, we find that r = -1 is a root and using polynomial division we get

r3+2r2鈭�19r鈭�20=(r+1)(r2+r鈭�20)=(r+1)(r+5)(r鈭�4)=0

03

General solution:

Now the roots of auxiliary equation arer1=鈭�1,r2=鈭�5 and . r3=4Therefore, a general solution to given equation is

y(x)=c1e鈭�x+c2e鈭�5x+c3e4x

Hence the final solution isy(x)=c1e鈭�x+c2e鈭�5x+c3e4x

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Most popular questions from this chapter

find a differential operator that annihilates the given function.

xe3xcos5x

Derive the system (7) in the special case when n=3. [Hint: To determine the last equation, require thatI.[yp]=g and use the fact that y1,y2, and satisfy the corresponding homogeneous equation.]

Find a general solution to the givenhomogeneous equation.

(D鈭�1)3(D鈭�2)(D2+D+1)(D2+6D+10)3[y]=0

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

(a) Because the ri鈥檚 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鈫掆垶 on the left-hand side to obtainC1 = 0.(b) Since C1 = 0, equation (33) becomes

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

and let x鈫掆垶 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(鈭�,-鈭�).

Use the annihilator method to show that ifa0=0anda1鈮�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+鈰�+B1x+B0}
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