Complete Solutions Manual For

Complete Solutions Manual For

Differential Equations with Boundary-Value Problems

NINTH EDITION

Dennis G. Zill Prepared by Roberto Martinez 1 / 4

Chapter 1 Introduction to Differential Equations 1.1 Definitions and Terminology 1.Second order; linear 2.Third order; nonlinear because of (dy/dx) 4 3.Fourth order; linear 4.Second order; nonlinear because of cos(r+u) 5.Second order; nonlinear because of (dy/dx) 2 or p

• + (dy/dx)

2 6.Second order; nonlinear because ofR 2 7.Third order; linear 8.Second order; nonlinear because of ˙x 2 9.Writing the differential equation in the formx(dy/dx) +y 2 = 1, we see that it is nonlinear inybecause ofy 2 . However, writing it in the form (y 2 −1)(dx/dy) +x= 0, we see that it is linear inx.

10.Writing the differential equation in the formu(dv/du)+(1+u)v=ue u we see that it is linear inv. However, writing it in the form (v+uv−ue u )(du/dv)+u= 0, we see that it is nonlinear inu.

11.Fromy=e −x/2 we obtainy ′ =− 1 2 e −x/2 . Then 2y ′ +y=−e −x/2 +e −x/2 = 0.

12.Fromy= 6 5 − 6 5 e −20t we obtaindy/dt= 24e −20t , so that dy dt

• 20y= 24e

−20t

• 20

θ 6 5 − 6 5 e −20t

= 24.

13.Fromy=e 3x cos 2xwe obtainy ′ = 3e 3x cos 2x−2e 3x sin 2xandy

′′

= 5e 3x cos 2x−12e 3x sin 2x, so thaty

′′

−6y ′

• 13y= 0.

1 2 / 4

2 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS

14.Fromy=−cosxln(secx+ tanx) we obtainy ′ =−1 + sinxln(secx+ tanx) and y

′′

= tanx+ cosxln(secx+ tanx). Theny

′′

+y= tanx.

15.The domain of the function, found by solvingx+2≥0, is [−2,∞). Fromy ′ = 1+2(x+2)

−1/2

we have (y−x)y ′ = (y−x)[1 + (2(x+ 2)

−1/2

] =y−x+ 2(y−x)(x+ 2)

−1/2

=y−x+ 2[x+ 4(x+ 2) 1/2 −x](x+ 2)

−1/2

=y−x+ 8(x+ 2) 1/2 (x+ 2)

−1/2

=y−x+ 8.An interval of definition for the solution of the differential equation is (−2,∞) becausey ′ is not defined atx=−2.

16.Since tanxis not defined forx=π/2 +nπ,nan integer, the domain ofy= 5 tan 5xis {x

5x6=π/2 +nπ} or{x

x6=π/10 +nπ/5}. Fromy ′ = 25 sec 2 5xwe have y ′ = 25(1 + tan 2 5x) = 25 + 25 tan 2 5x= 25 +y 2 .An interval of definition for the solution of the differential equation is (−π/10, π/10). Another interval is (π/10,3π/10), and so on.

17.The domain of the function is{x

4−x 2 6= 0}or{x

x6=−2 andx6= 2}. From y ′ = 2x/(4−x 2 ) 2 we have y ′ = 2x θ 1 4−x 2

2 = 2xy 2 .An interval of definition for the solution of the differential equation is (−2,2). Other intervals are (−∞,−2) and (2,∞).

18.The function isy= 1/ √ 1−sinx, whose domain is obtained from 1−sinx6= 0 or sinx6= 1.Thus, the domain is{x

x6=π/2 + 2nπ}. Fromy ′ =− 1 2 (1−sinx)

−3/2

(−cosx) we have 2y ′ = (1−sinx)

−3/2

cosx= [(1−sinx)

−1/2

] 3 cosx=y 3 cosx.An interval of definition for the solution of the differential equation is (π/2,5π/2). Another one is (5π/2,9π/2), and so on. 3 / 4

∞.∞ De×nitions and Terminolo}y 3 19.Writing ln(2X−1)−ln(X−1) =tand differentiating implicitly we obtain 2

2X−1

dX dt − 1 X−1 dX dt = 1 θ 2

2X−1

− 1 X−1

dX dt = 1

2X−2−2X+ 1

(2X−1) (X−1)

dX dt = 1 dX dt

=−(2X−1)(X−1) = (X−1)(1−2X).

– 4–2 2 4

t

– 4

–2 2 4 x Exponentiating both sides of the implicit solution we obtain

2X−1

X−1 =e t 2X−1 =Xe t −e t (e t −1) = (e t

−2)X

X= e t −1 e t −2 .Solvinge t −2 = 0 we gett= ln 2. Thus, the solution is defined on (−∞,ln 2) or on (ln 2,∞).The graph of the solution defined on (−∞,ln 2) is dashed, and the graph of the solution defined on (ln 2,∞) is solid.

20.Implicitly differentiating the solution, we obtain −2x 2 dy dx −4xy+ 2y dy dx = 0 −x 2 dy−2xy dx+y dy= 0 2xy dx+ (x 2 −y)dy= 0.Using the quadratic formula to solvey 2 −2x 2 y−1 = 0 fory, we gety= − 2x 2 ± √ 4x 4

• 4

· /2 =x 2 ± √ x 4

• 1 .

Thus, two explicit solutions arey1=x 2 + √ x 4

• 1 and

y2=x 2 − √ x 4

• 1 . Both solutions are defined on (−∞,∞).

The graph ofy1(x) is solid and the graph ofy2is dashed.

– 4–2 2 4

x

– 4

–2 2 4 y

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