nagle_differential_equations_ISM_Part9

nagle_differential_equations_ISM_Part9 - Chapter 2 12. We...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 2 12. We have 3 vdv 1- 4 v 2 = dx x Z 3 vdv 1- 4 v 2 = Z dx x - 3 8 Z du u = Z dx x ( u = 1- 4 v 2 , du =- 8 vdv ) - 3 8 ln 1- 4 v 2 = ln | x | + C 1 1- 4 v 2 = exp- 8 3 ln | x | + C 1 = Cx- 8 / 3 , where C = e C 1 is any nonzero constant. Separating variables, we lost constant solutions satisfying 1- 4 v 2 = 0 v = 1 2 , which can be included in the above formula by letting C = 0. Thus, v = 1- Cx- 8 / 3 2 , C arbitrary , is a general solution to the given equation. 14. Separating variables, we get dy 1 + y 2 = 3 x 2 dx Z dy 1 + y 2 = Z 3 x 2 dx arctan y = x 3 + C y = tan ( x 3 + C ) , where C is any constant. Since 1 + y 2 6 = 0, we did not lose any solution. 16. We rewrite the equation in the form x (1 + y 2 ) dx + e x 2 ydy = 0 , separate variables, and integrate. e- x 2 xdx =- ydy 1 + y 2 Z e- x 2 xdx =- Z ydy 1 + y 2 Z e- u du =- dv v ( u = x 2 , v = 1 + y 2 ) - e- u =- ln | v | + C ln ( 1 + y 2 )- e- x 2 = C is an implicit solution to the given equation. Solving for y yields y = p C 1 exp [exp (- x 2 )]- 1 , where C 1 = e C is any positive constant, 36 Exercises 2.2 18. Separating variables yields dy 1 + y 2 = tan xdx Z dy 1 + y 2 = Z tan xdx arctan y =- ln | cos x | + C. Since y (0) = 3, we have arctan 3 =- ln cos 0 + C = C C = 3 ....
View Full Document

Page1 / 5

nagle_differential_equations_ISM_Part9 - Chapter 2 12. We...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online