{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

s08week06

# s08week06 - Week 6,7 2.5 Wronskian 3.1 Intro Slope Fields...

This preview shows pages 1–4. Sign up to view the full content.

Week 6,7: 2.5 Wronskian 3.1 Intro, Slope Fields, verify solution 3.2 Separable DE 3.3 Exact DE 3.4 Linear Equations 3.6 mixing/cooling ————– Problem: Verify that the function y = c 1 x is a solution of y = y 2 x Solution: Compute y and check. y = c 1 ( 1 2 ) x - 1 2 . y 2 x = c 1 x 2 x = c 1 ( 1 2 ) x ( x ) 2 = c 1 ( 1 2 ) 1 x = y . —————–

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

View Full Document
2 Problem: Determine all values r so y = e rx is a solution to y - 4 y + 3 y = 0 . Main Step: to solve linear DE dy dx + p ( x ) y = q ( x ) multiply by the integral factor f = e R p dx and use d dx ( fy ) = f ( y + py ) on the left to get d dx ( fy ) = fq. Problem: Solve dy dx + 2 x (1 - x 2 ) y = 4 x, - 1 x 1 . Solution: Find integral factor, inside integral first: Z 2 x (1 - x 2 ) dx = - ln(1 - x 2 ) = ln ( (1 - x 2 ) - 1 ) (simplify!). ————– So e R 2 x ( 1 - x 2 ) dx = e ln ( (1 - x 2 ) - 1 ) = ( 1 - x 2 ) - 1 = 1 1 - x 2 = f.
3 Multiply by f = 1 1 - x 2 and use Main Property: 1 1 - x 2 dy dx + 2 x (1 - x 2 ) y = 4 x 1 - x 2 , d dx y 1 - x 2 = 4 x 1 - x 2 . Now integration gives y 1 - x 2 = ( - 2 ln (1 - x 2 )) + c, so y = (1 - x 2 ) ( - ln ((1 - x 2 ) 2 ) + c ) .

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

View Full Document
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