Problemas 2 5

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Exercises 2.5 Solutions by Substitutions

2. Letting y = ux we have

(x+ ux) dx + x{ u dx + x du) = 0

(1 + 2-u) dx + x du = 0

dx du-- h --- x 1 + 2u

= 0

In |x| + - In |1 + 2u\ = c

rr2 ( l + 2 | ) = C

x 1 + 2x y = ci.

3. Letting x = vy we liavc

vy{v dy + y dv) + (y - 2vy) dy = 0

vy2dv + y (y2 - 2v + l) dy = 0

vdv + rfy = ()

In

In \v 1|

x

(u - l )2 y

1

y

v - l

1

x / y - 1

+ In \y\ = c

+ ln y = c

(x - y )In \x - y \ - y = c(x - y ).

4. Letting x = vy wc have

y(v dy + ydv) 2(vy + y) dy = 0

y dv (t> + 2) dy = 0

dv _ dy ^ 0

v + 2 y

In \v+ 2| In \y\ = c

Inx

+ 2; - In \y| = c

x + 2y = ciy

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5. Letting y = ux wc have

(u?x2 + ux 2 j dx x 2(u dx -r x du) = 0

u2dx x du = 0

d x d u _9 u

X v>

In | . t | + = cu

In Irrl + = cV

yIn \x\ + x = cy.

6. Letting y - ux and using partial fractions, we have

( u2x 2 + ux 2'} dx + x 2 (u dx + x du) = 0

x2 (v? + 2it} dx + x 3du = 0

dx du + 7

- - -= 0x u(u+ 2)

In |jc| + ^ In | w[ ^ In \u+ 2| = c

ci

X \x

,2_

7. Letting y = ux wc have

x ry = c\ (y+ 2x).

(ux x) dx (ux+ x) (udx + x du) = 0

(u2 + l) dx+ x(u + 1) du = 0

dx u + 1 , + - s - du = 0

x u2 + 1

In jx| + ^ In (u2 + l) + t an-1 u = c

ln x 2 + 1^ + 2 tan-1 - = c

In fa;2 + y2) + 2 tan 1= c\ .' ' x

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(a; + 3ux) dx (3a- -+ux ) (u dx+ x du) = 0

('u2 l ) dx + x(u + 3) du = 0

dx it+ 3- - - h 7- - - rz- - - r r du = 0X ( u l ) ( l l + l )

In |.-r) + 2 In \u 1| In | u+ 1) = c

x(u l ) 2

Letting y = ux we have

u4- 1= Cl

(y - x f = c.i{y -+ x).


ux dx + (x+ \ fuX')(u dx+ x du) = 0

(x 2 + ) du + x u^'2dx = 0

1\ , dx(V 3 /2 + i U + * = 0V u) x

2w- 1/2 + In \u\ + In |x| = c

In \ y/x\ + In |xj = 2\ jx fy + c

y (ln \y\ - c)2 = 4x.


(uz + \ jx2 (ux )2 ) da: x (udx + x du) du = 0

\ /x2 u2x 2 dx x 2 du = 0

x\ j1 u2 dx x2 du 0, (x > 0)

dx du= 0

X V l - t i 2

In x sin- 1 u = c

sin 1u = In x+ ci

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S il l ' = 111 X + C2X

y . ,, N= sm(in x+ c'2 )

y = xs in (In x + C2 ).

See Problem 33 in this section for an analysis of the solution.

11. Letting y ux we have

(x 3 w3x 3) dx+ u2x (u dx+ x du) 0

dx + u2x du = 0

dx 9 ,h u du = 0

x

, , < 1 3In |x| + - u = co

3.x3 In |x| + y3 = cix3.

Using y (l) = 2 we find c\ = 8. T he solution of the initial- value problem is 3x:iIn |x| + y i 8x3.

12. Letting y ux we have

(x2 4- 2u?x 2)dx ux 2(udx-f xdu) = 0

x2(l + u2)dx ux 3 du = 0

dx u duX 1 + u2

In |x[ ^ ln(l -f u2) = c


x 2= C l

1 + u2

X 4

- ci(x 2 + y2).

Using y (1) = 1 we find ci 1/2. T he solution of the initial- value problem is 2x4 = y2 + x 2.

13. Letting y = ux we have

(x + uxeu) dx xe (u dx + x du) = 0

dx xeu du = 0

dxz z _ eu du = 0x

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In |a*| eu = c

In \x\ - eyf x = c.

Using y (l) = 0 wc find c= 1. The solution of the initial- value problem is In |.x[ = ey,lx 1.

Letting x = vy we have

y(v dy+ y dv) + vy(In vy In y 1) dy = 0

y dv + vIn v dy 0

dv dy+ = 0

vIn v yIn | In I'l.'H+ In | y| = c

yIn = ci.

Using y (l) e we find ci = e. T he solution of the initial- value problem is yInX\

= e.y

r- / 1 1 -o o . dw 3 3 . , oFrom y H y y and w = y we obtain |-w . A n integrating factor is x so that

x ' x ' dx x x

x3w = x 3 + c or t/3 = 1 + cx~A.

(IZUFrom y' y exy2 and w = y~ l we obtain I-w = ex. A n integrating factor is ex so that

dxf:xw ~ Ie 2x + c. or y~ l = \ ex + ce~x.

From y' + y = x y4 and w = we obtain 3u> = 3x. A n integr ating factor is e~ix so thatdxe~'ixw = xe~ ix + + c or ;ty_a = x + ^ -i- cejx .

From y y = y2 and w - y~ l we obtain ^ + 1+ w = 1. A n integr ating factor is

1 f*xex so that xexw = xex + exH- cor y-1 = 1 + + e~x.

x x

/ 1 1 o , _i , . dto 1 1rrom y - y = - y and w = y wc obtain + w = A n integrating factor is t so that

1 t-tw In i -|- c or y~ l - Inf + Wr iting this in the form - = hit + c. we see that the solution

t t y

can also be expressed in the form e^y = cit.

/ 2 2t a , _> . . dw 2t 21From y + 3 (1+ ^ */= 3 (1 + t2)y ai W = V W 0 am ~dt ~ l + f i W = T T f5 mtcgratmg

-| ^* X factor is 80 that 1 ^ 2 = + c or y~* = 1+ c (l + t2).

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/ 2 3 4 . _o , . dw 6 9 . . . . . .21. Prom y -- y = and w = y we obta in hw = ^ . A n integr ating factor is x so tha:

x x * ' dx x x z

x 6w = x5+ cor y~ 3 = x_1 + cx~b. If y(1) = ^ then c= ^ and y~ 3 = f x-1 + ^x - 6.

dtxi. 3 322. From i f + y = y~1//2 and w = y3/2 we obtain + - w = - . A n integrating factor is e3x-/2 so tha:

fix z zeix l 2w (?x>2 + c or y3/2 = 1 + ce~iX/'2. If y(0) = 4 then c = 7 and ,(/}/2 = 1 + 7e~^xt2.

23. Let u = x - f' + l so that du/dx = 1 + dy/dx. T hen ^ 1 = u2 or -xdu = dx. Tlni-dx 1 + t r

ta n-5 u = x + c or u = tan(x + c), and x + y+ 1 = t.an(x + c) or y tan(x + c) x 1.

24. Let u ~ x+ y so that du/dx = 1 + dy/dx. Then ~ 1 = - - --

or u du = dx. T hus Aw,2 = x + .dx uor u2 = 2x + ci. and (x+ y )2 = 2x+ c j.

25. Let u = x + y so that du/dx = 1 + dy/dx. Then ^ 1 = tan2u or cos2 udu = dx. Thu?\JfJU

^ + | s in2u = x + cor 2+ sin2u = 4x + ei, and 2(x + y)+ sin2(x + y) : 4x+ ci or 2y+ sin2(x+ y) =

2x + ci.

26. Let u x + y so that du/dx = 1 + dy/dx. T hen 1 = sin uo r--

du = dx. Multiplyingctcc 1 sm *?/1 sin

bv (1 sin if)/(1 sin u) we have - - - ~ du = dx or (sec2u secu taiiu)du = dx. Thu-cos^ u

tan usec u = x + cor tan(x + y) sec(x + y) = x + c.

27. Let u = y 2.x + 3 so that du/dx = dy/dx 2 . T hen ^ + 2 = 2 + v / or ~ t=du = dx. Thu-ax y w

2-y/w= x + c and 2^/y 2x + 3 = x + c.

Let u = y x + 5 so that du/dx =

e~u = x + c and ey~x+5 = x + c.

du28. Let u = y x + 5 so that du/dx = dy/dx 1. T hen - I- 1 = 1 + eu or e~udu = dx. Thi;.-

dx

ctu 129. Let u = x + y so that du/dx = 1 + dy/dx. Then - -- 1 cos u a n d - - - - :-- du. = dx. Now

fix 1+ cos u

1 1 cos u 1 COS u O---------------------------------------- = CSC u CSC ucot u1 + cos u 1 cos2u sin2u

so we have f(csc2u- c.sc ucot u)du = / dxand cot w+ csc u= x + c . Thus cot(x+y)+ csc(x+ y) - |

x + c. Setting x = 0 and y = tt/4 we obtain c = a/ 2 1 . The solution is

csc(x + y) - cot(x + y) = X+ V2 1.

30. Let u = 3x + 2yso that du/dx= 3+ 2dy/dx. T hen ^ = 3+ - = ~r~ and ^ du d.iy ' J l dx u + 2 u+ 2 5u+ 6

Now bv long divisionu + 2 1 4

^ +5u+ 6 5 25u + 30

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+ ~ ^ ) du = dx2bu + 30,

~o we have

Idand ^ In | 25?i + 30| = x+ c. T hus

1 4~(3x + 2y) -f- In | 75x + 50y + 30| = x + c.D 2 d

Setting x = 1 and y = 1 we obtain c = ^ In 95. T he solution is

1 4 4-(3a: + 2y)+ I n \ 7ox + 50y + 30| = x+ In 95o 25 2o

Jl'5y 5x+ 2 In |75.x + 50y + 30) = 2 In 95.

We wr ite the differential equation M(x , y)dx+ N(x , y)dy = 0 as dy/dx f (x , y) where

f u v) =/l ,W N(x. y) '

The function f i x , y)must necessarily be homogeneous of degree 0 when M and N are homogeneousjf degree a. Since M is homogeneous of degree a, M(tx 7ty) = taM(x ,y), and letting t = 1/x we

have

M ( l ,y /x ) = M(x, y) or M (x,y) = x aM (1,y /x ).

Thus

i y = f ( T = r f *M { l ,y /x ) M ( l , y /x ) fy \

dx x a N ( l ,y /x ) N ( l ,y /x ) \ x )

Rewrite (5:r2 2y2)dx x y dy = 0 as

^ = 5*2- 25,*dx

and divide by xy, so that

We then identify

dy x ^y- r = 5 - - 2 - .

dx y x

X / \ x - )XJ

(a) By inspection y = x and y = x are solutions of the differential equation and not members of

the family y ~ xsin (In a; + 02)-

(b) Letting x = 5 and y = 0 in sin- 1(y/x) = In x+ 0 2 we get sin- 10 = In 5 + c or c = In 5. T hensin- 1(y/x) = In xIn 5 = ln(,x/5). Because the range of the arcsine function is [7r /2, tt/2] we

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must have7T x -k

< ln - < 2 ~ 5 ~ 2

e-7r/2 < - < e*-/25

5e_ ^/2 < a: < 5ew/2.

The interval of definition of the solution is approximately

[1.04,24.05].

34. As x *oc. 0 and y *2x + 3. Now write (1 -I- ce6* )/( l ce6x) as (e~bx+ c)/(e~ 6x

Then, as x>oo, (?~fxT >0 and y>2x 3.

35. (a) The substitutions y yi + u and

dy _ clyi | du

dx dx dx

lead to

+ y- = P + Q(tli + u) + R(yi + u)2dx dx

= P + Qy\ -j- Ry'i + Qu+ 2y\ Ru+ R u2

or

~ ~ { Q + 2yi R )u = R u2.dx

This is a Bernoulli equation with n = 2 which can be reduced to the linear equation

div h (Q+ 2y\ R)tv R dx

bv the substitution w = u

(b) Identify P(x) = - 4/x 2, Q{x) = - 1/x , and R(x) = 1. Then ~ - f - + - )w = - 1. .- j

- ldw

dx

\x + cx~3

1 4

x x ,- l 2

. T hus, y ~ h u.x

integrating factor is a3 so that x^w = | x4+ e or u =

36. Wr ite the differential equation in the form x(y'/y) In x+ Inymid let u lny. Then du/dx = y' j

and the differential equation becomes x(du/dx) = ln x -j- u or du/dx u/x = (lnx )/x . which j

first- order and linear. A n integr ating factor is e~ J dx/a: = 1/x. so that (using integration by par:-

d_

dx

lnx uu ! = 7T and - =

x J r / x

1 lnx

x x+ c.

The solution is

Iny = 1 In x -r cx or y =X

37. Wr ite the differential equation as

- - v = 32vdx x

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; ::d let u = v2 or v = u1^. Thendv _ 1 _ xj 2 du

dx 2 d x '

.-.lid substituting into the differ ential equation, wc have1 i / o dlli 1 i / o - 1 / 9 d

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