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We have used various substitutions for yto transform the original equation into a new equation that we could solve. In some cases we must transform both xand yinto new variables, say,uand y. This is the situation for equations with linear coefficients—that is, equations of the form

(13) Aa1xb1yc1Bdx ^Aa2xb2yc2Bdy0 , y³

y0 y² x

2 1

20Ce^10x . yy²

y x 2 1

20Ce^10x . dy

dx 10y5x . 1

2 dy

dx 5y 5

2x ,

dy

/

^{dx 2y3 dy}

/

^dx,

yy². y³dy

dx5y² 5 2x .

n3, PAxB 5, and QAxB 5x

/

^2.

dy

dx5y 5

2xy³ . 1

/

^A1nB

1 1n

dy

dx PAxByQAxB . dy

dx ^A1nByⁿdy dx , yy¹ⁿ,

yⁿdy

dxPAxBy¹ⁿQAxB . yⁿ 72 Chapter 2 First-Order Differential Equations

Example 3

Solution

where the ai’s, bi’s, and ci’s are constants. We leave it as an exercise to show that when equation (13) can be put in the form which we solved via the substitution

Before considering the general case when let’s first look at the special situa-tion when Equation (13) then becomes

which can be rewritten in the form

This equation is homogeneous, so we can solve it using the method discussed earlier in this section.

The above discussion suggests the following procedure for solving (13). If then we seek a translation of axes of the form

where h and k are constants, that will change and change

Some elementary algebra shows that such a transformation exists if the system of equations

(14)

has a solution. This is ensured by the assumption , which is geometrically equiva-lent to assuming that the two lines described by the system (14) intersect. Now if satisfies (14), then the substitutions transform equation (13) into the homo-geneous equation

(15)

which we know how to solve.

Solve (16)

Since we will use the translation of axes ,

, where hand ksatisfy the system

Solving the above system for hand k gives Hence, we let and . Because dy dyand dxdu, substituting in equation (16) for xand yyields

dy

du 3(y

/

^u)

1(y

/

^{u) .}

A3uy_Bdu ^Auy_Bdy0

yy3

xu1 h1, k 3.

hk20 . 3hk60 , yyk

xuh a₁b₂ ^A3B A1B ^A1B A1Ba₂b₁,

A3xy6Bdx ^Axy2Bdy0 .

dy du

a₁ub₁y a₂ub₂y

a₁b₁Ay

/

^uB

a₂b₂Ay

/

^{uB ,}

xuh and yyk

Ah, kB a₁b₂a₂b₁

a1hb1kc10 , a2hb2kc20 a₂xb₂yc₂ into a₂ub₂y.

a₁xb₁yc₁ into a₁ub₁y

xuh and yYk ,

a₁b₂a₂b₁, dy

dx

a₁xb₁y a₂xb₂y

a₁b₁Ay

/

^xB

a₂b₂Ay

/

^{xB .}

Aa₁xb₁yBdx Aa₂xb₂yBdy0 , c₁c₂0.

a₁b₂a₂b₁, zaxby.

dy

/

^dxGAaxbyB,

a₁b₂a₂b₁,

Section 2.6 Substitutions and Transformations 73

Example 4

Solution

The last equation is homogeneous, so we let .Then and, sub-stituting for , we obtain

Separating variables gives

from which it follows that

When we substitute back in for z,u, andy, we find

This last equation gives an implicit solution to (16). ◆

Ay3B²2Ax1B Ay3B3Ax1B²C .

y²2uy3u²C , Ay

/

^uB22Ay

/

^uB3Cu² , z²2z3Cu² .

1

2 ln 0z²2z30 ln 0u0 C₁ ,

z^2z2z¹3dz ^{1udu ,}

zudz

du3z 1z . y

/

^u

dy

/

^duzuAdz

/

^duB,

zy

/

^u

74 Chapter 2 First-Order Differential Equations

In Problems 1–8, identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form

1.

2.

3.

4.

5.

6.

7.

8.

Use the method discussed under “Homogeneous Equa-tions” to solve Problems 9–16.

9.

10.

11.

12.

13.

14. dy

du u secAy

/

^uBy

u dx

dt x²t2t²x² tx

Ax²y²Bdx2xy dy0

Ay²xyBdxx²dy0

A3x²y²Bdx ^Axyx³y¹Bdy0

Axyy²Bdxx²dy0

Ay³uy²Bdu2u²y dy0

cosAxyBdysinAxyBdx

Aye^2xy³Bdxe^2xdy0

udyy du 2uy du

Atx2Bdx ^A3tx6Bdt0

dy

/

^dxy

/

^xx3y2

Ay4x1B²dxdy0

2tx dx ^At²x²Bdt0 y¿GAaxbyB.

15. 16.

Use the method discussed under “Equations of the Form

” to solve Problems 17–20.

17. 18.

19. 20.

Use the method discussed under “Bernoulli Equations”

to solve Problems 21–28.

21.

22.

23.

24.

25. 26.

27. 28. dy

dxy³xy0 dr

du r²2ru u²

dy

dxye^xy² dx

dt tx³x t 0 dy

dx y

x25Ax2By¹

/

²

dy dx 2y

x x²y² dy

dxye^2xy³ dy

dx y xx²y²

dy

/

^dxsinAxyB

dy

/

^{dx Axy5B2}

dy

/

^dxAxy2B2

dy

/

^{dx 2xy1}

dy

/

^dxGAaxbyB

dy

dx yAln yln x1B x

dy

dx x²y² 3xy

2.6 ^EXERCISES

Use the method discussed under “Equations with Linear Coefficients” to solve Problems 29–32.

29.

30.

31.

32.

In Problems 33–40, solve the equation given in:

33. Problem 1. 34. Problem 2.

35. Problem 3. 36. Problem 4.

37. Problem 5. 38. Problem 6.

39. Problem 7. 40. Problem 8.

41. Use the substitution to solve equa-tion (8).

42. Use the substitution to solve

43. (a) Show that the equation is

homogeneous if and only if [Hint:Let .]

(b) A function is called homogeneous of

order nif Show that the

equation

is homogeneous if are both

homogeneous of the same order.

44. Show that equation (13) reduces to an equation of the form

when [Hint: If , then

so that .]

45. Coupled Equations. In analyzing coupled equa-tions of the form

dx

dt axby , dy

dt axby ,

a₂ka₁ and b₂kb₁ a₂

/

^a1b2

/

^b1k,

a₁b₂a₂b₁ a₁b₂a₂b₁.

dy

dxGAaxbyB ,

MAx, yB and NAx, yB MAx, yBdxNAx, yBdy0

HAtx, tyBtⁿHAx, yB. HAx, yB

t1

/

^x

fAtx, tyB fAx, yB. dy

/

^{dxfAx, yB}

dy dx2y

x cosAy

/

^{x2B .}

yyx² yxy2

A2xy4Bdx Ax2y2Bdy0

A2xyBdx A4xy3Bdy0

Axy1Bdx Ayx5Bdy0

A3xy1Bdx Ax y3Bdy0

Section 2.6 Substitutions and Transformations 75

†Historical Footnote:Count Jacopo Riccati studied a particular case of this equation in 1724 during his investigation of curves whose radii of curvature depend only on the variable yand not the variable x.

where are constants, we may wish to determine the relationship between x and y rather than the individual solutions For this purpose, divide the first equation by the second to obtain

(17)

This new equation is homogeneous, so we can solve it via the substitution We refer to the solu-tions of (17) as integral curves.Determine the inte-gral curves for the system

46. Magnetic Field Lines. As described in Problem 20 of Exercises 1.3, the magnetic field lines of a dipole satisfy

Solve this equation and sketch several of these lines.

47. Riccati Equation. An equation of the form (18)

is called a generalized Riccati equation.^†

(a) If one solution — say, — of (18) is known, show that the substitution reduces (18) to a linear equation in y.

(b) Given that is a solution to

use the result of part (a) to find all the other solu-tions to this equation. (The particular solution can be found by inspection or by using a Taylor series method; see Section 8.1.) uAxBx

dy

dxx³AyxB²y x , uAxBx

yu1

/

^y

uAxB dy

dxPAxBy²QAxByRAxB dy

dx 3xy 2x²y² . dx

dt 2xy . dy

dt 4xy ,

yy

/

^x.

dy

dx axby axby .

xAtB, yAtB. a, b, a, and b

In this chapter we have discussed various types of first-order differential equations. The most important were the separable, linear, and exact equations. Their principal features and method of solution are outlined below.

Separable Equations: Separate the variables and integrate.

Linear Equations: . The integrating factor reduces

the equation to

Exact Equations: Solutions are given implicitly by . If , then is exact and Fis given by

or

When an equation is not separable, linear, or exact, it may be possible to find an integrat-ing factor or perform a substitution that will enable us to solve the equation.

Special Integrating Factors: is exact. If depends

only on x, then

is an integrating factor. If depends only on y, then

is an integrating factor.

Homogeneous Equations: . Let . Then

and the transformed equation in the variables yand xis separable.

Equations of the Form: . Let zax by.Then and the transformed equation in the variables zand xis separable.

Bernoulli Equations: . For or 1, let . Then

and the transformed equation in the variables yand xis linear.

Linear Coefficients: For , let

and , where hand ksatisfy

Then the transformed equation in the variables uand yis homogeneous.

a₂hb₂kc₂0 . a₁hb₁kc₁0 ,

yyk xuh

a₁b₂a₂b₁ Aa1xb1yc1Bdx ^Aa2xb2yc2Bdy0.

dy

/

^{dx A}1nByⁿAdy

/

^dxB,

yy¹ⁿ n0

dy

/

^dxPAxByQAxByⁿ

dz

/

^dxabAdy

/

^dxB,

dy

/

^dxGAaxbyB

dy

/

^dxyxAdy

/

^dxB,

yy

/

^x

dy

/

^dxGAy

/

^xB

m_AyBexpc^a0N

^/

^0xM0M

^/

^0ybdyd

A0N

/

^0x0M

/

^0yB

/

^M

m_AxBexpc^a0M

^/

^{0yN 0N}

^/

^0xbdxd

A0M

/

^0y 0N

/

^0xB

/

^N

MM dxMN dy0

F ^{N dyhAxB , where h¿AxBM 0x0 N dy .}

F ^{M dxgAyB , where g¿AyBN 0y0 M dx}

M dxN dy0

0N

/

^{0x 0}

M

/

^0y

FAx, yBC dFAx, yB0.

dAmyB

/

^dxmQ, so that mymQ dxC.

mexp

3

^PAxBdx

4

dy

/

^dxPAxByQAxB

dy

/

^dxgAxBpAyB.

76 Chapter 2 First-Order Differential Equations

Chapter Summary

In Problems 1–30, solve the equation.

1. 2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23. AyxBdx ^AxyBdy0

A2x 2y8Bdx ^Ax3y6Bdy0

Ay 2x1Bdx ^Axy4Bdy0

dy du y

u 4uy²

Ax²3y²Bdx2xy dy0 dy

dx ^A2xy1B² dy

du 2yy² dy

dxy tan xsin x0 dy

dx2 22xy3 dx

dt x

t1 t²2 dy

dx y

x x²sin 2x

Ay³4e^xyBdx ^A2e^x3y²Bdy0 dx

dt 1cos²AtxB

3^y1

/

² _A1x²2xyy²B¹4^dy0 3^{1 A}1x²2xyy²B¹4^dx

Ax²y²Bdx3xy dy0 dy

dx 2y

x 2x²y² t³y² dtt⁴y⁶ dy0 2xy³ dx A1x²Bdy0

3^{sinAxyBxy cosAxyB}4^dx3^1x2cosAxyB4^dy0 dy

dx 3y

x x²4x3

Ax²2y³Bdy ^A2xy3x²Bdx0

dy

dx 4y32x² dy

dx e^xy y1

In document Estudio sobre las familias migrantes y la incidencia en las relaciones escolares y familiares de los hijos, realizado en el sexto y séptimo año de educación general básica, de la escuela dos de agosto de la parroquia de ciano cantón Puyango provincia de L (página 71-76)