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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 249
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

dxdt+x+dydt=e4t,2x+d2ydt2=0

The solutions for the given linear system are xt=c1e-2t+c2et+29e4t and yt=-12c1e-2t-2c2et-136e4t+c3.

See the step by step solution

Step by Step Solution

Step 1: General form

Elimination Procedure for 2 × 2 Systems

To find a general solution for the system

L1x+L2y=f1,L3x+L4y=f2,

Where L1,L2,L3, and L4 are polynomials in D=ddt

  1. Make sure that the system is written in operator form.
  2. Eliminate one of the variables, say, y, and solve the resulting equation for x(t). If the system is degenerating, stop! A separate analysis is required to determine whether or not there are solutions.

  1. (Shortcut) If possible, use the system to derive an equation that involves y(t) but not its derivatives. [Otherwise, go to step (d).] Substitute the found expression for x(t) into this equation to get a formula for y(t). The expressions for x(t), and y(t) give the desired general solution.

  1. Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants----in fact, twice as many as needed.]

  1. Remove the extra constants by substituting the expressions for x(t) and y(t) into one or both of the equations in the system. Write the expressions for x(t) and y(t) in terms of the remaining constants.

Step 2: Evaluate the given equation

Given that,

dxdt+x+dydt=e4t … (1)

2x+d2ydt2=0 … (2)

Let us rewrite the system in operator form,

D+1x+Dy=e4t … (3)

2x+D2y=0 … (4)

Multiply D on both sides of equation (3) then subtract equation (3) and (4) together one gets,

DD+1x+D2y-2x+D2y=De4tD2+D-2x=4e4tD2+D-2x=4e4t5

Since the auxiliary equation to the corresponding homogeneous equation is r2+r-2=0. The roots are r=1 and r=-2 .

Then, the homogeneous solution of u is;

xht=c1e-2t+c2et … (6)

Let us take the undetermined coefficients and assume that,

xpt=Ae4t … (7)

Now find the derivate the equation (7).

Dxpt=4Ae4tD2xpt=16Ae4t

Step 3: Substitution method

Substitute the derivation in equation (5).

D2+D-2Ae4t=4e4t16Ae4t+4Ae4t-2Ae4t=4e4t18Ae4t=4e4t

Now, equalize the like terms.

18A=4A=418A=29

So, xpt=29e4t … (8)

Use equations (6) and (8) to get,

xt=xht+xptxt=c1e-2t+c2et+29e4t9

Now, take equation (3).

D+1x+Dy=e4tDy=e4t-D+1x=e4t-D+1c1e-2t+c2et+29e4t=e4t+2c1e-2t-c2et-89e4t-c1e-2t-c2et-29e4t=c1e-2t-2c2et-19e4tyt=-12c1e-2t-2c2et-136e4t+c3

Thus, the solutions for the given linear system are xt=c1e-2t+c2et+29e4t and yt=-12c1e-2t-2c2et-136e4t+c3.

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