# HW7 - ma293hw7solns.nb 1 MA293 Homework 7 Solutions 11 yH4L...

• Notes
• 2

This preview shows pages 1–2. Sign up to view the full content.

MA293 Homework 7 Solutions 11. y H 4 L - 8 y H 3 L + 16 y H 2 L = 0 r 4 - 8 r 3 + 16 r 2 = 0 fl r 2 H r 2 - 8 r + 16 L = 0 fl r 2 H r - 4 L 2 = 0 So r = 0, 0, 4, 4 gives y H x L = c 1 + c 2 x + c 3 4 x + c 4 x 4 x 23. y '' - 6 y ' + 25 y = 0 ; y H 0 L = 3 ; y ' H 0 L = 1 r 2 - 6 r + 25 = 0 r 1,2 = 6 è!!!!!!!!!!!!!!!!!!!!!!! 36 - 25 * 4 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 2 = 3 4 i So our general solution is y H x L = ‰ 3 x H c 1 cos H 4 x L + c 2 sin H 4 x LL y H 0 L = c 1 = 3 y ' H x L = ‰ 3 x H 4 c 2 cos H 4 x L - 12 sin H 4 x LL + 3 3 x H 3 cos H 4 x L + c 2 sin H 4 x LL y ' H 0 L = 9 + 4 c 2 = 1 fl c 2 = - 2 y H x L = ‰ 3 x H 3 cos H 4 x L - 2 sin H 4 x LL 28. 2 y ''' - y '' - 5 y ' - 2 y = 0 2 r 3 - r 2 - 5 r - 2 = 0 One root is r = 2 H r - 2 L H 2 r 2 + 3 r + 1 L = 0 H r - 2 L H r + 1 L H r + 1 ê 2 L y H x L = c 1 2 x + c 2 x + c 3 x ê 2 34. 3 y ''' - 2 y '' + 12 y ' - 8 y = 0 ; y 1 = ‰ 2 ÅÅÅÅ 3 x 3 r 3 - 2 r 2 + 12 r - 8 = 0 One root is r = 2 ê 3 H r + 2 ê 3 L H 3 r 2 + 12 L = 0 H r + 2 ê 3 L H r + 2 I L H r - 2 I L = 0 y H x L = c 1 2 ÅÅÅÅ 3 x + c 2 cos H 2 x L + c 3 sin H 2 x L 49. y H 4 L = y ''' + y '' + y ' + 2 y ma293hw7solns.nb 1

This preview has intentionally blurred sections. Sign up to view the full version.

r 4 - r 3 - r 2 - r - 2 = 0 H r - 2 L H r + 1 L H r 2 + 1 L = 0 r = 2, - 1, i , - i y H x L = c 1 2 x + c 2 - x + c 3 cos H x L + c 4 sin H x L Now apply initial conditions: y H 0 L = 0 = c 1 + c 2 + c 3 y ' H 0 L = 0 = 2 c 1 - c 2 + c 4 y '' H 0 L = 0 = 4 c 1 + c 2 - c 3 y ''' H 0 L = 30 = 8 c 1 - c 2 - c 4 Solving this system of equations yields: c 1 = 2, c 2 = - 5, c 3 = 3, c 4 = - 9 Solution: y H x L = 2 2 x - 5 - x + 3 cos H x L - 9 sin H x L 54. x 3 y ''' + 6 x 2 y '' + 4 xy ' = 0 dy ÅÅÅÅÅÅÅ dx = dy ÅÅÅÅÅÅÅ dv 1 ÅÅÅÅ x d 2 y ÅÅÅÅÅÅÅÅÅÅ dx 2 = - dy ÅÅÅÅÅÅÅ
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: d v 2 d 3 y ÅÅÅÅÅÅÅÅÅÅ d x 3 = 1 ÅÅÅÅÅ x 3 I d 3 y ÅÅÅÅÅÅÅÅ d v 3-3 d 2 y ÅÅÅÅÅÅÅÅ d v 2 + 2 d y ÅÅÅÅÅÅ d v M Substituting this into the differential equation yields: y ''' H v L-3 y '' H v L + 2 y ' H v L + 6 y '' H v L-6 y ' H v L + 4 y ' H v L = y ''' H v L + 3 y '' H v L = 0. r 3 + 3 r 2 = r 2 H r + 3 L y H v L = c 1 + c 2 v + c 3 ‰-3 v So y H x L = c 1 + c 2 ln H x L + c 3 ‰-3 ln H v L = c 1 + c 2 ln H x L + c 3 x-3 ma293hw7solns.nb 2...
View Full 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