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Linear Algebra With Applications
Found in: Page 442
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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Short Answer

Solve the differential equation f'''(t)-f''(t)-4f'(t)+4f(t)=0and find all the real solution of the differential equation.

The solution is ft=C1et+C2e2t+C3e-2t.

See the step by step solution

Step by Step Solution

Step1: Definition of characteristic polynomial

Consider the linear differential operator


The characteristic polynomial of T is defined as


Step2: Determination of the solution

The characteristic polynomial of the operator as follows.


Then the characteristic polynomial is as follows.

role="math" localid="1660798972108" PTλ=λ3λ24λ+4

Solve the characteristic polynomial and find the roots as follows.

λ3λ24λ+4=0λ2λ14λ1=0 λ24λ1=0λ+2λ2λ1=0

Simplify further as follows.

λ1=0 λ=1λ+2=0 λ=2

Simplify further as follows.

λ2=0 λ=2

Therefore, the roots of the characteristic polynomials are 1,2 and 2.

Step3: Conclusion of the solution

Since, the roots of the characteristic equation is different real numbers.

The exponential functions et,e2t and e-2t form a basis of the kernel of T .

Hence, they form a basis of the solution space of the homogenous differential equation is Tf=0.

Thus, the general solution of the differential equation f'''tf''t4f't+4f't=0 is ft=C1et+C2e2t+C3e-2t.

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