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: 11

Answer

1) We show that really $y(0)=4$ i.e. that $y$ satisfies the given initial condition. 2) We find $y'$ and put $y$ and $y'$ into the given differential equation to show that $y$ really is its' solution. by 1) and 2) we verified the particular solution.

Work Step by Step

1) Show that $y(0) = 4$. Indeed $$y(0)=4e^{-6\times0^2} = 4e^0 = 4\times 1 = 4.$$ 2) Show that $y$ really is a solution to this equation. First we will find $y'$: $$y'=(4e^{-6x^2})'=4(e^{-6x^2})' = 4e^{-6x^2}(-6x^2)' = -24e^{-6x^2}2x = -48xe^{-6x^2}.$$ Now we put $y$ and $y'$ into the given differential equation The Left side is: $$y'=-48xe^{-6x^2}.$$ The Right side is: $$-12xy = -12x(4e^{-6x^2}) = -48xe^{-6x^2}.$$ Since both sides are equal to each other we see that $y$ really is a solution to this differential equation. 1) and 2) say that $y$ is verified as a particular solution to this 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.