95_pdfsam_math 54 differential equation solutions odd

95_pdfsam_math 54 differential equation solutions odd - xy...

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

View Full Document Right Arrow Icon
Review Problems given in Section 2.4. First we integrate M ( x, y ) with respect to x to get F ( x, y )= Z ( 2 xy 3 x 2 ) dx + g ( y ) F ( x, y )= x 2 y x 3 + g ( y ) . (2.22) To determine g ( y ) take the partial derivative with respect to y of both sides and substitute N ( x, y )for ∂F ( x, y ) /∂y to obtain N = x 2 2 y 3 = x 2 + g 0 ( y ) . Solving for g 0 ( y ) yields g 0 ( y )= 2 y 3 . Since the choice of the constant of integration is arbitrary we will take g ( y )= y 2 . Hence, from equation (2.22) we have F ( x, y )= x 2 y x 3 + y 2 and the solution to the di±erential equation is given implicitly by x 2 y x 3 + y 2 = C . 5. In this problem, M ( x, y )=s in( xy )+ xy cos( xy ) ,N ( x, y )=1+ x 2 cos( xy ) . We check the equation for exactness: ∂M ∂y =[ x cos( xy )] + [ x cos( xy ) xy sin(
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: xy ) x ] = 2 x cos( xy ) − x 2 y sin( xy ) , ∂N ∂x = 0 + [2 x cos( xy ) − x 2 sin( xy ) y ] = 2 x cos( xy ) − x 2 y sin( xy ) . Therefore, the equation is exact. So, we use the method discussed in Section 2.4 and obtain F ( x, y ) = Z N ( x, y ) dy = Z ± 1 + x 2 cos( xy ) ² dy = y + x sin( xy ) + h ( x ) ⇒ ∂F ∂x = sin( xy ) + x cos( xy ) y + h ( x ) = M ( x, y ) = sin( xy ) + xy cos( xy ) ⇒ h ( x ) = 0 ⇒ h ( x ) ≡ , and a general solution is given implicitly by y + x sin( xy ) = c . 91...
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 Berkeley.

Ask a homework question - tutors are online