33_pdfsam_math 54 differential equation solutions odd

# 33_pdfsam_math 54 differential equation solutions odd -...

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Exercises 2.2 x ( 1+ Ce t ) = t x = t t . Note: When C is replaced by K , this answer can also be written as x = Ke t / ( t 1). Further we observe that since we divide by x x 2 = x (1 x ), then x 0and x 1area lso solutions. Allowing K to be zero gives x 0, but no choice for K will give x 1, so we list this as a separate solution. 15. To separate variables, we move the term containin dx to the right-hand side of the equation and divide both sides of the result by y . This yields y 1 dy = ye cos x sin xdx y 2 dy = e cos x sin xdx. Integrating the last equation, we obtain Z y 2 dy = Z ( e cos x sin x ) dx ⇒− y 1 + C = Z e u du ( u =cos x ) 1 y + C = e u = e cos x y = 1 C e cos x , where C is an arbitrary constant. 17. First we ±nd a general solution to the equation. Separating variables and integrating, we get dy dx = x 3 (1
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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 Berkeley.

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