sol - 4. (a) Note that this equation is a linear first...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 4. (a) Note that this equation is a linear first order differential equation. dy − 2y = x dx ˆ I (x) = exp( −2dx) = e−2x (1) (2) dy −2x e − 2ye−2x = xe−2x dx (3) Dx (e−2x y ) = xe−2x (4) Take integrals both sides: ˆ e−2x y = xe−2x dx Now, we solve this equation for y. 1 Use integration by parts: u = x, du = dx, dv = e−2x dx, v = − 2 e−2x . ˆ ˆ 1 1 1 −2x 1 −2x + e−2x dx = − xe−2x − e−2x + C xe dx = − xe 2 2 2 4 (5) (6) 1 1 e−2x y = − xe−2x − e−2x + C 2 4 (7) 1 1 y = − x − + Ce2x 2 4 (8) 11 1 = − − + Ce2 24 (9) Use y (1) = 1 to find C : C= 7 4 e2 (10) Now, use this to find the general solution: 1 1 7 y = − x − + 2 e2x 2 4 4e 4. (b) Note that x0 = 1, y0 = 1, and h = 1.4−1 2 = 0.2 since we’re using two steps of Euler’s method. g (x, y ) = dy = x + 2y dx 1 x1 = 1.2, y1 = y0 + g (x0 , y0 )h = 1 + g (1, 1) · 0.2 = 1 + (1 + 2) = 1.6 5 x2 = 1.4, y2 = y1 + g (x1 , y1 )h = 1.6 + g (1.2, 1.6) · 0.2 = 1.6 + 0.2(1.2 + 3.2) = 2.48 Therefore, y (1.4) (11) (12) (13) (14) 2.48. 5. (a) Let y the amount of lead in the tank at time t, then y is increasing in a constant rate and decreasing in a rate of proportional to its current amount (y ). dy y = (3 × 2) − , y (0) = 0 dt 2 1 (15) Note that this is a seperable differential equation. Rearranging terms leads to: dt dy = 30 − y 5 (16) Take integrals both side: ˆ dy = 30 − y ˆ − ln |30 − y | = dt 5 (17) t +C 5 (18) Use y (0) = 0, then C = − ln 30 (19) Plug this back into equation (18) and solve for y , then we get the general solution: y = 30 − 30e−t/5 (20) lim 30 − 30e−t/5 = 30 (21) 5. (b) t→∞ Therefore, the amount of lead in the tank approaches to 30 as time goes to infinity. 5. (c) y (t) = 10 (22) 30 − 30e−t/5 = 10 (23) 3 et/5 = ln( ) 2 (24) t 3 = ln( ) 5 2 (25) 3 t = 5 ln( ) 2 (26) Therefore, lead concentration in the tank reaches 1 gram per liter when t = 5 ln( 3 ). 2 2 ...
View Full Document

Page1 / 2

sol - 4. (a) Note that this equation is a linear first...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online