# week05 - dy dx + p ( x ) y = q ( x ) multiply by the...

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

Week 5: 2.5 Wronskian 3.1 Intro, Slope Fields, verify solution 3.2 Separable DE 3.4 Linear Equations ————– 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: Inﬂamable substance, temp T 0 = 50 (F), placed in hot oven, temp T m = 450 (F). After 20 min substance temp T = 150 . Find temp at 40 min. If substance ignites at 350, ﬁnd time of combustion. Solution: Newton’s Law of cooling: T = T ( t ) temp at time t, dT dt = - k ( T - T m ) , T m = 450 , t in min. ————– dT dt = - k ( T - T m ) , T m = 450 , T (0) = 50 , T (20) = 150 , ﬁnd T (40) and t c so T ( t c ) = 350 . Method: separation. dT T - T m = - k dt, ( T ± = T m ). integral: ln | T - T m | = - k t + c, T m = 450 . Initial Data: ln (450 - T ) = - k t + c, T (0) = 50 , T (20) = 150 . When t = 0 , ln 400 = c, e c = 400 . So 450 - T = e - kt e c = 400 e - kt , and T = T ( t ) = 450 - 400 ( e - k ) t . Next, when t = 20 , 150 = 450 - 400 ( e - k ) 20 , so - 300 = - 400 ( e - k ) 20 , e - k = ( 3 4 ) 1 20 , and T ( t ) = 450 - 400 ( 3 4 ) t 20 . Finally, T ( t ) = 450 - 400 ( 3 4 ) t 20 , gives T (40) = 225 , and T ( t c ) = 350 gives t c = 96 . 4 minutes. (why?) ————
3 Main Step: to solve linear DE

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.

Unformatted text preview: 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. 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 . 4 Now integration gives y 1-x 2 = (-2 ln (1-x 2 )) + c, so y = (1-x 2 ) (-ln ((1-x 2 ) 2 ) + c ) . Notice that we can check this instance of the Main Property directly (using the product rule)....
View Full Document

## This note was uploaded on 02/29/2008 for the course MATH 205 taught by Professor Zhang during the Spring '08 term at Lehigh University .

### Page1 / 4

week05 - dy dx + p ( x ) y = q ( x ) multiply by the...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online