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Q37E

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Linear Algebra With Applications
Found in: Page 7
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

Find the functionf(t) of the form f(t)=ae3t+be2t such that f(0)=1and f'(0)=4.

The function of the form f(t)=ae3t+be2t such that f(0)=1 and f'(0)=4 is f(t)=2e3t-e2t.

See the step by step solution

Step by Step Solution

Step 1: Consider the points and substitute these in the standard equation

Consider the function f(t)=ae3t+be2t. Put f(0)=1 in f(t)=ae3t+be2t

f0=ae30+be20 1=a+b

Step 2: Consider the derivative of the polynomial

Take the derivative of the polynomial f(t)=ae3t+be2t and put f'(0)=4 into its derivative.

f'=t=3ae3t+2be2tf'0=3ae3×0+2be2×0 4=3a+2b

Step 3: Solve the above equations.

Arrange the equations obtained in above steps into the matrix form.

1132ab=14

Upon solving the values of a,b are obtained as a=2,b=-1

Substitute these values in the polynomial equation.

role="math" localid="1659340662448" f(t)=ae3t+be2tf(t)=2e3t-1e2t

The function ft of the form f(t)=ae3t+be2t such that f(0)=1 and f'(0)=4 is f(t)=2e3t-e2t.

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