66_pdfsam_math 54 differential equation solutions odd

# 66_pdfsam_math 54 differential equation solutions odd -...

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

Chapter 2 Integrating M ( x, y ) with respect to x yields F ( x, y )= Z M ( x, y ) dx = Z (cos x cos y +2 x ) dx =c o s y Z cos xdx + Z 2 xdx =s in x cos y + x 2 + g ( y ) . To fnd g ( y ), we compute the partial derivative oF F ( x, y ) with respect to y and compare the result with N ( x, y ). ∂F ∂y = ∂y ± sin x cos y + x 2 + g ( y ) ² = sin x sin y + g 0 ( y )= (sin x sin y +2 y ) g 0 ( y )= 2 y g ( y )= Z ( 2 y ) dy = y 2 . (We take the integration constant C = 0.) ThereFore, F ( x, y )=s in x cos y + x 2 y 2 = c is a general solution to the given equation. 13. In this equation, the variables are y and t , M ( y,t )= t/y , N ( y,t )=1+ln y .S ince ∂M ∂t = ∂t ³ t y ´ = 1 y and ∂N ∂y = ∂y (1 + ln y )=
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Integrating M ( y, t ) with respect to y , we get F ( y, t ) = Z t y dy = t ln | y | + g ( t ) = t ln y + g ( t ) . (±rom N ( y, t ) = 1 + ln y we conclude that y > 0.) ThereFore, ∂F ∂t = ∂ ∂t [ t ln y + g ( t )] = ln y + g ( t ) = 1 + ln y ⇒ g ( t ) = 1 ⇒ g ( t ) = t ⇒ F ( y, t ) = t ln y + t, and a general solution is given by t ln y + t = c (or, explicitly, t = c/ (ln y + 1)). 62...
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