nagle_differential_equations_ISM_Part9

# nagle_differential_equations_ISM_Part9 - Chapter 2 12 We...

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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 = π 3 ....
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nagle_differential_equations_ISM_Part9 - Chapter 2 12 We...

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