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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.

D2u+Dv=2,4u+Dv=6

The solutions for the given linear system are ut=c1e-2t+c2e2t+1 and vt=2c1e-2t-2c2e2t+2t+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

a. Make sure that the system is written in operator form.

b. 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.

c. (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.

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

e. 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,

D2u+Dv=21

4u+Dv=62

Then subtract equation (1) and (2) together one gets,

D2u+Dv-4u+Dv=2-6D2-4u=-4D2-4u=-43

Since the auxiliary equation to the corresponding homogeneous equation is r2-4=0. The roots are r=-2 and r=2.

Then, the homogeneous solution of u is;

uht=c1e-2t+c2e2t4

Let us take the undetermined coefficients and assume that,

upt=15

Step 3: Substitution method

Use equations (4) and (5) to get,

ut=uht+uptut=c1e-2t+c2e2t+16

Now, take equation (2).

4u+Dv=6Dv=6-4c1e-2t+c2e2t+1Dv=-4c1e-2t-4c2e2t+2vt=2c1e-2t-2c2e2t+2t+c3

Thus, the solutions for the given linear system are ut=c1e-2t+c2e2t+1 and vt=2c1e-2t-2c2e2t+2t+c3.

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