Diff Eqs 3 solutions

# Diff Eqs 3 solutions - dx = x t 2 x 0 = 1 dt The general...

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Differential Equations 3 – Answers and some Solutions 1. 2 t x dt dx - = , ( 29 1 0 = x The general form of the d.e. for Euler’s method is ( 29 x t f dt dx , = In this example, the function, ( 29 2 , t x x t f - = The initial conditions are written as 0 0 = t , 1 0 = w where 0 t is the initial time and 0 w is the solution at this time. Euler’s method is defined by: ( 29 w w hf t w i i i i + = + 1 , and i takes the values 0, 1, 2, … in turn So the solution to the equation is built up in steps. Start with i = 0. This is the initial time at which the solution is known. The new time is always h t t i i + = + 1 and the solution at this time is given by ( 29 w w hf t w i i i i + = + 1 , So with i = 0, the new time is 1 . 0 0 1 = + = h t t and ( 29 0 0 0 1 , w t hf w w + = ( 29 ( 29 1 . 1 0 1 1 . 0 1 1 , 0 1 . 0 1 2 = - + = + = f ( 29 1 , 0 f means the function ( 29 2 , t x x t f - = evaluated using 0 for the time, t , and 1 for the solution, x . Now move on to the next step. i = 1, the new time is 2 . 0 1 2 = + = h t t , and the solution at this time is ( 29 ( 29 ( 29 209 . 1 1 . 0 1 . 1 1 . 0 1 . 1 1 . 1 , 1 . 0 1 . 0 1 . 1 , 2 1 1 1 2 = - + = + = + = f w t hf w w One more step is required to obtain the solution at time 0.3.

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i = 2, the new time is 3 . 0 2 3 = + = h t t , and the solution at this time is ( 29 ( 29 ( 29 3259 . 1 2 . 0 209 . 1 1 . 0 209 . 1 209 . 1 , 2 . 0 1 . 0 209 . 1 , 2 2 2 2 3 = - + = + = + = f w t hf w w The solution approximated at the various time steps using Euler’s method is t solution, w 0 1 (the initial condition) 0.1 1.1 0.2 1.209 0.3 1.3259 2. The solution approximated at the various time steps using Euler’s method is t approximate solution, w exact solution, x 1 2 (the initial condition) 2 1.25 2 2.05 1.5 2.1 2.1667 1.75 2.25 2.32143 2 2.428571 2.5 Exact solution: with some rearrangement, the d.e. takes the general form of a linear equation and can be solved with the use of an integrating factor. i.e. the equation can be written as 2 = + t x dt dx The integrating factor is t e e t dt t = = ln 1 The general solution is found to be t c t x + = where c is an arbitrary constant. Using the initial condition, ( 29 2 1 = x , the constant is evaluated as 1 = c . The particular solution of the d.e. is therefore t t x 1 + = For comparison with the approximate solution obtained using Euler’s method, the exact solution at the various times has been included in the table above.
3. The general solutions of these homogeneous equations are: (a) ( 29 x B x A e y x 3 sin 3 cos + = where A and B are arbitrary constants. (b) 0 2 2 2 = + - y dx dy dx y d The first step is to write the auxiliary equation. 0 1 2 2 = + - m m We want to find the roots of this quadratic equation since the form of the solution of the d.e. depends on the type of roots. ( 29 (

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## This note was uploaded on 04/25/2010 for the course MATH 201 taught by Professor Any during the Spring '10 term at Westminster UT.

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Diff Eqs 3 solutions - dx = x t 2 x 0 = 1 dt The general...

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