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 - Page 403: 32


The particular solution that passes through the point $(3,4)$ is $$y=\sqrt{2x^2-2}.$$

Work Step by Step

Here the solution is given in its' implicit form $2x^2-y^2=C$ and it contains an arbitrary constant $C$. To find required $C$ we have to demand that the solution passes through the point $(3,4)$ i.e. when we put $x=3$ we get $y=4$: $$2\times3^2-4^2=C\Rightarrow C=2.$$ Now we have $$2x^2-y^2=2.$$ From here we will find $y$ explicitly: $$2x^2-y^2=2\Rightarrow y^2=2x^2-2\Rightarrow y=\pm\sqrt{2x^2-2}.$$ We have to throw away the solution with the $"-"$ sign because when we put $x=3$ we get $$y=-\sqrt{2\times3^2-2}=-\sqrt{16}=-4$$ which we did not want. When we take the $"+"$ sign we have for $x=3$ $$y=\sqrt{2\times3^2-2}=\sqrt{16}=4,$$ and that is exactly what we wanted. So the particular solution that passes through the point $(3,4)$ is $$y=\sqrt{2x^2-2}.$$
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.