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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 μ ( t ) ] μ ( t ) = exp ±Z t 1 1 s ds ² = exp h (ln s ) ³ ³ 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 µ ds + 1 6 ² = 1 t " ´ 1 2 s 21 3 s µ ³ ³ ³ ³ t 1 + 1 6 # = 1 t ± 1 2 t 21 3 t1 2 + 1 3 + 1 6 ² = 1 2 t1 3...
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This note was uploaded on 06/12/2008 for the course MATH 20D taught by Professor Mohanty during the Spring '06 term at UCSD.
 Spring '06
 Mohanty
 Math

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