101_pdfsam_math 54 differential equation solutions odd

101_pdfsam_math 54 differential equation solutions odd -...

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Review Problems 27. This equation has linear coefficients. Thus we make a substitution x = u + h , y = v + k with h and k satisfying ( 3 h k 5=0 h k +1 = 0 h =3 , k =4 . With this substitution, (3 u v ) du +( u v ) dv =0 dv du = 3 u v u v = 3 ( v/u ) 1 ( ) z = v u ,v = uz, v 0 = z + uz 0 z + u dz du = 3 z 1 z u dz du = 3 z 1 z z = z 2 3 z 1 z 1 z 2 3 dz = du u Z z 1 z 2 3 dz = Z du u . We use partial fractions to Fnd the integral in the left-hand side. Namely, z 1 z 2 3 = A z 3 + B z + 3 ,A = 1 2 1 2 3 ,B = 1 2 + 1 2 3 . Therefore, integration yields A ln ± ± ± z 3 ± ± ± + B ln ± ± ± z + 3 ± ± ± = ln | u | + C 1 ² z 3 ³ 1 1 / 3 ² z + 3 ³ 1+1 / 3 u 2 = C ² v u 3 ³ 1 1 / 3 ² v + u 3 ³ 1+1 / 3 = C ´ v 2 3 u 2 ) v + u 3 v u 3 1 / 3 = C µ ( y 4) 2 3( x 3) 2
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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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