49_pdfsam_math 54 differential equation solutions odd

49_pdfsam_math 54 differential equation solutions odd - ( t...

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Exercises 2.3 Thus, the solution is y = xe x x, on the interval (0 , ) . Note: This interval is the largest interval containing the initial value x = 1 in which P ( x )and Q ( x ) are continuous. 19. In this problem, t is the independent variable and x is the dependent variable. One can notice that the left-hand side is the derivative of xt 3 with respect to t . Indeed, using product rule for diFerentiation, we get d dt ( xt 3 ) = dx dt t 3 + x d ( t 3 ) dt = t 3 dx dt +3 t 2 x. Thus the equation becomes d dt ( xt 3 ) = t xt 3 = Z tdt = t 2 2 + C x = t 3 ± t 2 2 + C ² = 1 2 t + C t 3 . (Of course, one could divide the given equation by t 3 to get standard form, conclude that P ( t )=3 /t , ±nd that
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Unformatted text preview: ( t ) = t 3 , multiply by t 3 back, and come up with the original equation.) We now use the initial condition, x (2) = 0, to nd C . 0 = x (2) = 1 2(2) + C 2 3 1 4 + C 8 = 0 C = 2 . Hence, the solution is x = 1 / (2 t ) 2 / ( t 3 ). 21. Putting the equation in standard form yields dy dx + sin x cos x y = 2 x cos x dy dx + (tan x ) y = 2 x cos x. Therefore, P ( x ) = tan x and so ( x ) = exp Z tan x dx = exp ( ln | cos x | ) = | cos x | 1 . 45...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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