91_pdfsam_math 54 differential equation solutions odd

91_pdfsam_math 54 differential equation solutions odd - t =...

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Exercises 2.6 du dx =( u 1) 2 4=( u 3)( u +1) Z du ( u 3)( u +1) = Z dx . To integrate the left-hand side, we use partial fractions: 1 ( u 3)( u +1) = 1 4 ± 1 u 3 1 u +1 ² . Thus 1 4 (ln | u 3 |− ln | u +1 | )= x + C 1 ln ³ ³ ³ ³ u 3 u +1 ³ ³ ³ ³ =4 x + C 2 u 3 u +1 = Ce 4 x u = Ce 4 x +3 1 Ce 4 x y =4 x + Ce 4 x +3 1 Ce 4 x , (2.21) where C 6 = 0 is an arbitrary constant. Separating variables, we lost the constant solutions u 3and u ≡− 1, that is, y =4 x +3and y =4 x 1. While y =4 x + 3 can be obtained from (2.21) by setting C =0 ,theso lut ion y =4 x 1 is not included in (2.21). Therefore, a general solution to the given equation is y =4 x + Ce 4 x +3 1 Ce 4 x and y =4 x 1
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Unformatted text preview: t = u + h and x = v + k with h and k satisfying ( h + k + 2 = 0 3 h k 6 = 0 h = 1 , k = 3 . As dt = du and dx = dv , the substitution yields ( u + v ) dv + (3 u v ) du = 0 du dv = u + v 3 u v = ( u/v ) + 1 3( u/v ) 1 . With z = u/v , we have u = vz , u = z + vz , and the equation becomes z + v dz dv = z + 1 3 z 1 v dz dv = 3 z 2 + 1 3 z 1 3 z 1 3 z 2 + 1 dz = 1 v dv Z 3 z 1 3 z 2 + 1 dz = Z 1 v dv 87...
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