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Differential Equations

SEPARATION OF VARIABLES

Graham S McDonald

A Tutorial Module for learning the techniqueof separation of variables

q Table of contentsq Begin Tutorial

c 2004 g.s.mcdonald@salford.ac.uk

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Table of contents1. Theory

2. Exercises

3. Answers

4. Standard integrals

5. Tips on using solutions

Full worked solutions

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Section 1: Th eory 3

1. Theor yIf one can re-arrange an ordin ary differential eq uation in to the fo llow-

ing standard form:

dydx

= f (x)g(y),

then the solution may be found by the technique of SEPARATIONOF VARIABLES :

dy

g(y)=

f (x) dx .

This result is obtained by dividing the standard form by g(y), andthen integrating both sides with respect to x.

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Section 2: Ex ercises 4

2. Exerci ses

Click on Exercise links for full worked solutions (there a re 16 exer-cises in total)

Exercise 1 .

Find the gen eral solu tion of dydx

= 3 x2e−y and the particul ar soluti on

that satises the condition y(0) = 1

Exercise 2 .

Find the gen eral solu tion of dydx

=yx

Exercise 3 .

Solve the equationdydx

=y + 1x −1

given the boun dary con dition: y = 1at x = 0

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Section 2: Ex ercises 7

Exercise 1 2.

Find the gen eral solu tion of 1

y

dy

dx=

x

x2 + 1

Exercise 1 3.

Solvedydx

=y

x(x + 1)and nd th e particu lar solut ion when y(1) = 3

Exercise 1 4.

Find the gen eral solu tion of sec x ·dydx

= sec 2 y

Exercise 1 5.

Find the gen eral solu tion of cosec3xdydx

= cos 2 y

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Section 2: Ex ercises 8

Exercise 1 6.

Find the gen eral solu tion of (1

−x2)

dy

dx+ x(y

−a) = 0 , where a is

a constant

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Section 3: An swers 9

3. Answe rs

1. General solu tion is y = ln( x3

+ A) , and part icular solution isy = ln( x3 + e) ,

2. General solu tion is y = kx ,

3. General solu tion is y + 1 = k(x −1) , and pa rticular solutionis y = −2x + 1 ,

4. General solu tion is y 3

3 = x 2

2 + C , and particul ar soluti on isy3 = 3x 2

2 + 1 ,

5. General solu tion is y = −ln −12 e2x −C , and pa rticular

solution is y = −ln 3−e 2 x

2 ,

6. General solu tion is ex

= ky(x + 1) ,

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Section 3: An swers 10

7. General solu tion is y2 − 1

4 sin2y = x 2

2 + 2 x + ln x + C ,

8. General solu tion is sin y = e−x 2 + A

, and part icular solution issin y = e−x 2

,

9. General solu tion is y(1 + x2)12 = k , and particular solution is

y(1 + x2)12 = 2 ,

10. General solu tion is ta n−1 y = ln x + C , and pa rticular solutionis tan −1 y = ln x + π

4 ,

11. General solu tion is y

−1 = kx2(y + 1 ) ,

12. General solu tion is y2 = k(x2 + 1) ,

13. General solu tion is y = kxx +1 , and par ticular so lution is y = 6x

x +1 ,

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Section 3: An swers 11

14. General solu tion is 2y + sin 2 y = 4 sin x + C ,

15. General solu tion is t an y = −cos x +13 cos

3

x + C ,

16. General solu tion is y −a = k(1 −x2)12 .

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Section 4: St andard int egrals 13

f (x) f (x) dx f (x) f (x) dx1

a 2 + x 2

1a tan −

1 xa

1a 2 −x 2

12a ln

a + xa −x (0 < |x|<a )

(a > 0) 1x 2 −a 2

12a ln x −a

x + a (|x| > a> 0)

1

√ a 2 −x 2 sin−1 x

a1

√ a 2 + x 2 lnx + √ a 2 + x 2

a (a > 0)

(−a < x < a ) 1√ x 2 −a 2 ln x + √ x 2 −a 2

a (x>a > 0)

√a2 −x2 a 2

2 sin−1 xa √a2 + x2 a 2

2 sinh−1 xa + x √ a 2 + x 2

a 2

+ x √ a 2 −x 2

a 2√x2 −a2 a 2

2 −cosh−1 xa + x √ x 2−a 2

a 2

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Section 5: Ti ps on usin g solutio ns 14

5. Tips o n usin g solut ions

q When looki ng at th e THEO RY, AN SWERS , INTEG RALS, orTIPS pages, use the Back button (at t he botto m of the page) toreturn to the exercises.

q Use the solu tions inte lligently. For exa mple, th ey can help you getstarted on an exercise, or they can allow you to check whether yourintermediate results are correct.

q Try to make less use of the f ull solut ions as you work your waythrough the Tutorial.

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Solutions to e xercises 15

Full work ed solu tionsExercise 1.

This is of the form dydx

= f (x)g(y) , where f (x) = 3 x2 and

g(y) = e−y , so we can separate the variables and then integrate,

i.e. ey dy = 3x2dx i.e. ey = x3 + A

(where A = arbitrary constant).

i.e. y = ln( x3 + A) : General solution

Particular so lution : y(x) = 1 w hen x = 0 i.e. e1 = 0 3 + A

i.e. A = e and y = ln( x3 + e) .

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Solutions to e xercises 16

Exercise 2.

This is of the formdy

dx

= f (x)g(y) , where f (x) = 1x and

g(y) = y, so we can separate the variables and then integrate,

dyy

= dxx

i.e. ln y = ln x + C = ln x + ln k (ln k = C = constant)

i.e. ln y

−ln x = ln k

i.e. ln(y/x ) = ln ki.e. y = kx .

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Solutions to e xercises 17

Exercise 3.Find the general solution rst. Then apply the boundary conditionto get the particular solution.

Equation is of the form:dydx

= f (x)g(y), where f (x) = 1x −1

g(y) = y + 1so separate variables and integrate.

i.e. dyy + 1

= dxx −1

i.e. ln(y + 1) = ln( x

−1) + C

= ln( x −1) + ln k (k = arbitrary constant)

i.e. ln(y + 1) −ln(x −1) = ln k

i.e. lny + 1x −1 = ln k

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Solutions to e xercises 18

i.e.y + 1x

−1

= k

i.e. y + 1 = k(x −1) (general solution)

Now determine k for particular solution with y(0) = 1.

x = 0y = 1 gives 1 + 1 = k(0 −1)

i.e. 2 = −ki.e. k =

−2

Particular solution: y + 1 = −2(x −1) i.e. y = −2x + 1 .

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S l i i 19

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Solutions to e xercises 19

Exercise 4.Use separation of variables to nd the general solution rst.

y2dy = x dx i.e.y3

3 =x2

2 + C

(general solu tion )

Particular so lution with y = 1 , x = 0 :13 = 0 + C i.e. C =

13

i.e. y3 = 3x 2

2 + 1 .

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S l ti t i 20

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Solutions to e xercises 20

Exercise 5.General solution rst then nd particular solution.

Write equation as: dydx = e2x ey (≡f (x)g(y))

Separate variables

and integrat e: dyey = e2x dx

i.e. −e−y = 12 e2x + C

i.e. e−y = −12 e2x −C

i.e. −y = ln −12 e2x −C

i.e. y = −ln −12 e2x −C .

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Solutions to e xercises 21

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Solutions to e xercises 21

Particular so lution : x = 0y = 0 gives 0 = −ln −1

2 −C

i.e. −12 −C = 1

i.e. C = −32

∴

y = −ln3

−e 2 x

2 .

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Solutions to e xercises 22

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Solutions to e xercises 22

Exercise 6.

Separate variables and integrate:

xx + 1

dx = dyy

Numerator and denominator of same degree in x: reduce degree of

numerator using long division.i.e. x

x +1 = x +1 −1x +1 = x +1

x +1 − 1x +1 = 1 − 1

x +1

i.e. 1 − 1x +1 dx = dy

y

i.e. x −ln(x + 1) = ln y + ln k (ln k = constant of integration)i.e. x = ln( x + 1) + ln y + ln k

= ln[ ky(x + 1)]

i.e. ex

= ky(x + 1) . General solution. Return to E xercise 6Toc Back

Solutions to e xercises 23

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Solutions to e xercises 23

Exercise 7.

Separate variables and integrate:

i.e. sin2 ydy = (x + 1) 2

xdx

i.e.

12

(1

−cos2y)dy =

x2 + 2x + 1

xdx

i.e.12 dy −

12 cos2ydy = x + 2 +

1x

dx

i.e.12

y −12 ·

12

sin2y =12

x2 + 2 x + l n x + C .

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Solutions to e xercises 24

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Solutions to e xercises 24

Exercise 8.General solution rst.

Separate variables: i.e. dytan y = −2x dx

Integrate: i.e. cot y dy = −2 xdx

i.e. ln(sin y) = −2 · x 2

2 + A

i.e. ln(sin y) = −x2 + A

i.e. sin y = e−x 2 + A

Note: cos ysin y

dy is of form f (y)f (y)

dy = ln [f (y)] + C

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Solutions to e xercises 25

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Solutions to e xercises 25

Particular so lution : y = π2 when x = 0

gives sin π2 = eA

i.e. 1 = eA

i.e. A = 0

∴ Required solution is sin y = e−x 2 .

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Solutions to e xercises 26

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Solutions to e xercises 26

Exercise 9.Separate variables and integrate:

(1 + x2) dydx = −xy

i.e. dyy

= − x1 + x2 dx

i.e. dy

y = −1

2 2x

1 + x2 dx[compare with f (x )

f (x ) dx]

i.e. ln y = −12 ln(1 + x2) + ln k (ln k = const ant)

i.e. ln y + ln(1 + x2) 12 = ln k

i.e. ln y(1 + x2)12 = ln k

i.e. y(1 + x2)12 = k, (general solution).

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Solutions to e xercises 27

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Particular so lution

y(0) = 2, i .e. y(x) = 2 when x = 0i.e. 2(1 + 0)

12 = k

i.e. k = 2i.e. y(1 + x2)

12 = 2 .

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Solutions to e xercises 28

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Exercise 1 0.

dy

y2 + 1 = dxx

Standard integral: dy1 + y2 = ta n−1 y + C

i.e. tan −1 y = ln x + C . General solution.

Particular solution with y = 1 when x = 1:

tan π4 = 1 ∴ tan −1(1) = π

4 , while ln 1 = 0 ( i.e. 1 = e0)

∴

π4 = 0 + C i.e. C =

π4

Particular solution is: tan −1 y = ln x + π4 .

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Solutions to e xercises 30

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i.e. Ay −1 + By +1 dy = d x

x

i.e. 1

2 1

y−1 −1

y +1dy =

d x

x

i.e. 12 [ln(y −1) −ln(y + 1 )] = ln x + ln k

i.e. ln(y −1) −ln(y + 1) −2 ln x = 2 ln k

i.e. ln y−1(y +1) x 2 = 2 ln k

i.e. y −1 = k x2(y + 1), ( k = k2 = constant) .

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Solutions to e xercises 31

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Exercise 1 2.

dyy =

xx2 + 1 dx =

12

2xx2 + 1 dx

Note : f (x)f (x)

dx = ln [f (x)] + A

i.e. ln y =12 ln x

2

+ 1 + C i.e.

12

ln y2 =12

ln x2 + 1 + C {get same coefficients to

allow log manipulations }i.e.

12 ln

y2

x2 + 1 = C

i.e.y2

x2 + 1= e2C

i.e. y2 = k x2 + 1 , (where k = e2C = constant) .

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Solutions to e xercises 33

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i.e. ln y −ln x + ln ( x + 1) = ln k (ln k = C = consta nt)

i.e. lny(x + 1)

x = ln k

i.e.y(x + 1)

x= k

i.e. y = kxx + 1

. General solu tion.

Particular solution with y(1) = 3:

x = 1, y = 3 gives 3 =k

1+1

i.e. k = 6

i.e. y = 6xx +1 .

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Solutions to e xercises 34

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Exercise 1 4.

dy

sec2 y = dx

sec x

i.e. cos2 y dy = cos x dx

i.e. 1 + cos 2y2

dy = cos x dx

i.e.y2

+12 ·

12

sin2y = sin x + C

i.e. 2y + sin 2 y = 4 sin x + C

(where C = 4 C = constant) .

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Solutions to e xercises 35

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Exercise 1 5.

i.e. dy

cos2 y = dx

cosec3x

= sin3 x dx

=

sin2 x

·sin x dx

= (1 −cos2 x) ·sin x dx

= sin x dx − cos2 x ·sin x dx

set u = cos x , sodudx

= −sin x

and cos 2 x ·sin x dx = −u2du

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Solutions to e xercises 36

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LHS is stand ard integ ral

sec2 y dy = tan y + A .

This gives, tan y = −cos x − −cos 3 x3 + C

i.e. tan y =

−cos x + cos 3 x

3+ C .

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Solutions to e xercises 37

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Exercise 1 6.

i.e. (1

−x2) dy

dx =

−x(y

−a)

i.e. dyy−a = − x1−x 2 dx

i.e. dyy−a = + 1

2 −2x1−x 2 dx [compare R HS integr al with f (x )

f (x ) dx]

i.e. ln(y −a) =12 ln(1 −x

2

) + ln ki.e. ln(y −a) −ln(1 −x2)

12 = ln k

i.e. ln y−a

(1

−x 2 )

12

= ln k

∴ y −a = k(1 −x2)12 .

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