ordinary differential equations separation variables
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8/8/2019 Ordinary Differential Equations Separation Variables
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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
q Theory q Answers q Integrals q Tips
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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 .
Return to E xercise 5
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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 .
Return to E xercise 7
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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 .
Return to E xercise 8
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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 .
Return to E xercise 9
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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 .
Return to E xercise 10
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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) .
Return to E xercise 11
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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 .
Return to E xercise 13
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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 .
Return to E xercise 15
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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 .
Return to E xercise 16
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