This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: (b) y t = y x + y 2 y t 2 Solution: secondorder, nonlinear, PDE (c) 2 y t 2 = 2 y e t + y x t 3 Solution: secondorder, linear, PDE 3. [ 10 points ] Solve the following initial value problem: 3 dy dt =3 t y + 31 t y (1) = 1 6 , t > Solution: Rewrite the equation as dy dt + 1 t y = 11 3 t Compute the integrating factor as [2 pts for setup, 2 pts for correct ( t ) ] ( t ) = exp Z t 1 1 s ds = exp h (ln s ) t 1 i = exp [ln t ] = t Use the integrating factor to compute the solution [2 pts for setup, 2 pts for correct y ( t ) up to a constant, 2 pts for the correct constant (to meet initial condition)] y ( t ) = 1 t Z t 1 s 11 3 s ds + 1 6 = 1 t " 1 2 s 21 3 s t 1 + 1 6 # = 1 t 1 2 t 21 3 t1 2 + 1 3 + 1 6 = 1 2 t1 3...
View Full
Document
 Spring '06
 Mohanty
 Math

Click to edit the document details