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chapter2.page07

# chapter2.page07 - y = e x 2 x sin y 3 y 2 3 Solution Write...

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2. SEPARABLE EQUATIONS 7 Now open Mathcad window, click any blank area, type [Ctrl][I] to get the antiderivative operator Z d type t^2 *e^{-3t} - 4 in the integrant place holder (again using space bar to exit the power mode) and t in the other place holder, then type [Ctrl][.], click any area outside the box. You will get Z t 2 e - 3 t - 4 dt = - 1 3 t 2 e - 3 t - 2 9 te - 3 t - 2 27 e - 3 t - 4 t + C (4) Put (3) and (4) together, we have ln | I ( t ) | = - 1 3 t 2 e - 3 t - 2 9 te - 3 t - 2 27 e - 3 t - 4 t + C Solve for I ( t ), we finally arrive at I ( t ) = e - 1 3 t 2 e - 3 t - 2 9 te - 3 t - 2 27 e - 3 t - 4 t + C = Ce - 1 3 t 2 e - 3 t - 2 9 te - 3 t - 2 27 e - 3 t - 4 t a The next example shows that sometimes we can find explicit for- mula for the solution of an separable ODE . Example
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Unformatted text preview: y = e x +2 x sin( y )+3 y 2 +3 Solution Write y as dy dx , multiply both sides of the equation by (sin( y ) + 3 y 2 + 3) dy and take antiderivative, we have Z sin( y ) + 3 y 2 + 3 dy = Z e x + 2 xdx Now using Mathcad or compute by hand, we ﬁnd Z sin( y ) + 3 y 2 + 3 dy =-cos( y ) + y 3 + 3 y + C and Z e x + 2 xdx = e x + x 2 + C So we have an implicitly deﬁned solution y ( x ) by the equation-cos( y ) + y 3 + 3 y = e x + x 2 + C Here, we can’t ﬁnd an explicit formula for y ( t ) . a Practice...
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