87_pdfsam_math 54 differential equation solutions odd

87_pdfsam_math 54 differential equation solutions odd -...

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Exercises 2.6 separate variables, and integrate. 1 dt dx =( t +5) 2 dt dx =1 ( t +5) 2 =(1 t 5)(1 + t +5)= ( t +4)( t +6) dt ( t +4)( t +6) = dx Z dt ( t +4)( t +6) = Z dx 1 2 Z ± 1 t +4 1 t +6 ² dt = Z dx ln ³ ³ ³ ³ t +4 t +6 ³ ³ ³ ³ = 2 x + C 1 ln ³ ³ ³ ³ x y +4 x y +6 ³ ³ ³ ³ = 2 x + C 1 x y +6 x y +4 = C 2 e 2 x 1+ 2 x y +4 = C 2 e 2 x y = x +4+ 2 Ce 2 x +1 . Also, the solution t +4 0 y = x +4 has been lost in separation variables. 21. This is a Bernoulli equation with n = 2. So, we make a substitution u = y 1 n = y 1 .W e have y = u 1 , y 0 = u 2 u 0 , and the equation becomes 1 u 2 du dx + 1 ux = x 2 u 2 du dx 1 x u = x 2 . The last equation is a linear equation with P ( x )= 1 /x
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Unformatted text preview: linear equations, we nd an integrating factor ( x ) = 1 /x and multiply the equation by ( x ) to get 1 x du dx 1 x 2 u = x d dx 1 x u = x 1 x u = Z ( x ) dx = 1 2 x 2 + C 1 u = 1 2 x 3 + C 1 x y = 1 x 3 / 2 + C 1 x = 2 Cx x 3 . Also, y 0 is a solution which was lost when we multiplied the equation by u 2 (in terms of y , divided by y 2 ) to obtain a linear equation. 83...
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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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