{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

63_pdfsam_math 54 differential equation solutions odd

# 63_pdfsam_math 54 differential equation solutions odd - as...

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

Exercises 2.4 Therefore, the temperature at noon (when t = 12 and j = 3) is T 3 (12) = C 3 e 12 = e 12 e 9 3 k =0 ( 1) 3 k e k = 1 e 1 + e 2 e 3 0 . 718 = 71 . 8 F . At 5 p.m. (when t = 17 and j = 8), we find T 8 (17) = 1 C 8 e 17 = 1 e 17 e 9 8 k =0 ( 1) 8 k e k = 8 k =1 ( 1) k +1 e k = e 1 · 1 ( e 1 ) 8 1 + e 1 0 . 269 = 26 . 9 F . EXERCISES 2.4: Exact Equations, page 65 1. In this equation, M ( x, y ) = x 2 y + x 4 cos x and N ( x, y ) = x 3 . Taking partial derivatives, we obtain ∂M ∂y = ∂y ( x 2 y + x 4 ) = x 2 = 3 x 2 = ∂N ∂x . Therefore, according to Theorem 2 on page 61 of the text, the equation is not exact. Rewriting the equation in the form dy dx = x 2 y + x 4 cos x x 3 = 1 x y + x cos x, (2.14) we conclude that it is not separable because the right-hand side in (2.14) cannot be factored
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: as p ( x ) q ( y ). We also see that the equation is linear with y as the dependent variable. 3. Here M ( x, y ) = ye xy + 2 x , N ( x, y ) = xe xy − 2 y . Thus ∂M ∂y = ∂ ∂y ( ye xy + 2 x ) = e xy + y ∂ ∂y ( e xy ) = e xy + ye xy x = e xy (1 + yx ) , ∂N ∂x = ∂ ∂x ( xe xy − 2 y ) = e xy + x ∂ ∂x ( e xy ) = e xy + xe xy y = e xy (1 + xy ) , ∂M/∂y = ∂N/∂x , and the equation is exact. We write the equation in the form dy dx = − ye xy + 2 x xe xy − 2 y 59...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online