Diff Eqs 3 solutions

Diff Eqs 3 solutions -...

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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.
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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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Diff Eqs 3 solutions -...

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