99_pdfsam_math 54 differential equation solutions odd

99_pdfsam_math 54 differential equation solutions odd - ( 2...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Review Problems Next we take the partial derivative of F with respect to y and substitute 2 x 3 y for ∂F/∂y : 2 x 3 y =2 x 3 y + g 0 ( y ) . Thus g 0 ( y ) = 0 and since the choice of the constant of integration is not important, we will take g ( y ) 0. Hence, we have F ( x, y )= x 1 + x 3 y 2 and the implicit solution to the diFerential equation is x 1 + x 3 y 2 = C. Solving for y 2 yields y 2 = x 2 + Cx 3 . ±inally we check to see if any solutions were lost in the process. We multiplied by the integrating factor µ ( x )= x 4 so we check x 0. This is also a solution to the original equation. 21. This equation has linear coefficients. Therefore, we are looking for a substitution x = u + h and y = v + k with h and k satisfying ( 2 h + k 1=0 h + k 4=0 h =1 , k =3 . So, x = u +1( dx = du )and y = v +3( dy = dv ), and the equation becomes
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ( 2 u + v ) du + ( u + v ) dv = 0 dv du = 2 u v u + v = 2 ( v/u ) 1 + ( v/u ) . With z = v/u , we have v = z + uz , and so z + u dz du = 2 z 1 + z u dz du = 2 z 1 + z z = z 2 2 z + 2 1 + z z + 1 z 2 + 2 z 2 dz = du u Z 1 + z z 2 + 2 z 2 dz = Z du u 1 2 ln z 2 + 2 z 2 = ln | u | + C 1 z 2 + 2 z 2 ) u 2 = C 2 . Back substitution, z = v/u = ( y 3) / ( x 1), yields v 2 + 2 uv 2 u 2 = C 2 ( y 3) 2 + 2( x 1)( y 3) 2( x 1) 2 = C 2 y 2 8 y 2 x 2 2 x + 2 xy = C. 95...
View Full Document

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.

Ask a homework question - tutors are online