{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

383_pdfsam_math 54 differential equation solutions odd

# 383_pdfsam_math 54 differential equation solutions odd -...

This preview shows page 1. Sign up to view the full content.

Exercises 6.4 Therefore, we have y p ( x ) = v 1 ( x ) + v 2 ( x ) cos x + v 3 ( x ) sin x = ln(sec x ) + cos 2 x + sin 2 x sin x ln(sec x + tan x ) y p ( x ) = ln(sec x ) sin x ln(sec x + tan x ) + 1 . Since y 1 is a solution to the homogeneous equation, we may choose y p ( x ) = ln(sec x ) sin x ln(sec x + tan x ) . Note: We left the absolute value signs off ln(sec x ) and ln(sec x + tan x ) because of the stated domain: 0 < x < π/ 2. 7. First, we divide the differential equation by x 3 to obtain the standard form y 3 x 1 y + 6 x 2 y 6 x 3 y = x 4 , x > 0 , from which we see that g ( x ) = x 4 . Given that { x, x 2 , x 3 } is a fundamental solution set for the corresponding homogeneous equation, we are looking for a particular solution of the form
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern