{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

9.8HW-solutions

# 9.8HW-solutions - -x h y ψ y = ∂ψ ∂y = x h y...

This preview shows pages 1–4. Sign up to view the full content.

MATH 2930: HW Solutions: Assigned September 8, 2010 Prepared by Alex Moore (problem #10 graded) Section 2.6 3.) (3 x 2 - 2 xy + 2)d x + (6 y 2 - x 2 + 3)d y = 0 ∂M ∂y = (3 x 2 - 2 xy + 2) ∂y = - 2 x ∂N ∂x = (6 y 2 - x 2 + 3) ∂x = - 2 x ∂M ∂y = ∂N ∂x This is an exact equation. We can use M to solve for the function: ψ = Z M d x = Z (3 x 2 - 2 xy + 2)d x = x 3 - x 2 y + 2 x + h ( y ) ψ y = ∂ψ ∂y = x 2 + h 0 ( y ) Comparing this to: N = (6 y 2 - x 2 + 3) h 0 ( y ) = 6 y 2 + 3 h ( y ) = 2 y 3 + 3 y + C The ﬁnal answer becomes: x 3 - x 2 y + 2 x + 2 y 3 + 3 y = C 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
10.) ( y/x + 6 x )d x + (ln x - 2)d y = 0 ∂M ∂y = ( y/x + 6 x ) ∂y = 1 /x ∂N ∂x = (ln x - 2) ∂x = 1 /x ∂M ∂y = ∂N ∂x This is an exact equation. We can use M to solve for the function: ψ = Z M d x = Z ( y/x + 6 x )d x = y ln x + 3 x 2 + h ( y ) ψ y = ∂ψ ∂y = ln x + h 0 ( y ) Comparing this to: N = ln x - 2 h 0 ( y ) = - 2 h ( y ) = - 2 y + C The ﬁnal answer becomes: y ln x + 3 x 2 - 2 y = C 14.) Note that the equation is in the form: M ( x,y )d x - N ( x,y )d y = 0 with a minus sign instead of a plus. (9 x 2 + y - 1)d x - (4 y - x )d y = 0 ∂M ∂y = (9 x 2 + y - 1) ∂y = 1 ∂N ∂x = - (4 y - x ) ∂x = 1 2
∂M ∂y = ∂N ∂x This is an exact equation. We can use M to solve for the function: ψ = Z M d x = Z (9 x 2 + y - 1)d x = 3 x 3 + xy

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: -x + h ( y ) ψ y = ∂ψ ∂y = x + h ( y ) Comparing this to: N =-4 y + x h ( y ) =-4 y ⇒ h ( y ) = 2 y 2 + C The ﬁnal answer becomes: 3 x 3 + xy-x-2 y 2 = C 16.) ( ye 2 xy + x )d x + ( bxe 2 xy )d y = 0 ∂M ∂y = ∂ ( ye 2 xy ) ∂y = e 2 xy + 2 xye 2 xy ∂N ∂x = ∂ ( bxe 2 xy ) ∂y = be 2 xy + 2 bxye 2 xy Equation is exact if b = 1. Using M to solve (and integrating by parts): ψ = Z M d x = Z ( ye 2 xy + x )d x = 1 2 e 2 xy + 1 2 x 2 + h ( y ) ψ y = ∂ψ ∂y = xe 2 xy + h ( y ) Comparing this to: N = xe 2 xy h ( y ) = 0 ⇒ h ( y ) = C 3 The answer becomes: 1 2 e 2 xy + 1 2 x 2 = C We can also multiply both sides by 2 and change 2 C → C to get the answer in the form: e 2 xy + x 2 = C 4...
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

9.8HW-solutions - -x h y ψ y = ∂ψ ∂y = x h y...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online