Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 6 - Differential Equations - 6.1 Exercises: 6

Answer

It is a solution of the differential equation.

Work Step by Step

Find the first and second derivatives. $y=C_1 e^{-x} cos{x} + C_2 e^{-x} sin{x}$ $y= e^{-x}(C_1cos(x) +C_2 sin(x))$ $y'= -e^{-x} ( C_1cos(x) + C_2 sin(x) ) + e^{-x} (-C_1 sin(x) +C_2 cos(x))$ $y'= e^{-x} [(C_2 -C_1)cos(x) - (C_1+C_2)sin(x)]$ $y''= -e^{-x}[(C_2-C_1)cos(x) - (C_1+C_2) sin(x)] + e^{-x}[-(C_2 - C_1)sin(x) - (C_1 + C_2) cos(x)] $ $y''= 2e^{-x}(C_1 sin(x) - C_2cos(x))$ Plug back into the differential equation $ 2e^{-x}((C_1 sin(x) - C_2cos(x)) + 2e^{-x} [(C_2 -C_1)cos(x) - (C_1+C_2)sin(x)] + 2e^{-x}(C_1cos(x) +C_2 sin(x)) =0 $ Simplify $2e^{-x}(C_1sin(x) -C_2cos(x) +C_2cos(x) -C_1 cos(x) -C_1sin(x) -C_2sin(x) + C_1cos(x) +C_2sin(x))= 0$ All terms cancel out and $0=0$ So, it is a solution to the differential equation.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.