SOLVED DIFFERENTIAL EQUATIONS

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Differential equation

A differential equation is an equation involving an unknown function and its derivatives. Also it

is an equation involving differential coefficients.

Examples

𝑑𝑦

𝑑𝑥= 3𝑥 + 5

𝑤ℎ𝑒𝑟𝑒: 𝑦 = 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒; 𝑥 = 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒

𝑑2𝑧

𝑑𝑥2+ 5

𝑑𝑧

𝑑𝑥+ 𝑎𝑧 = 0

𝑤ℎ𝑒𝑟𝑒: 𝑧 = 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒; 𝑥 = 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒

510xd

yd4

3

3

y

dx

dy


𝑒𝑦𝑑2𝑦

𝑑𝑥2+ 5 (

𝑑𝑦

𝑑𝑥)

2

= 1


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7𝑑3𝑦

𝑑𝑥3+ (𝑠𝑖𝑛𝑥)

𝑑2𝑦

𝑑𝑥2+ 9𝑥𝑦 = 0


𝑑2𝑧

𝑑𝑥2+ 5𝑥 (

𝑑𝑧

𝑑𝑥)

4

+ 3𝑧 = 𝑠𝑒𝑐𝑥

(𝑑4𝑦

𝑑𝑥4)

3

+ 4𝑦 (𝑑𝑦

𝑑𝑥)

5

+ 𝑦3 (𝑑𝑦

𝑑𝑥)

2

= 12𝑥


𝑑𝑦

𝑑𝑥= 𝑥 +

𝑘

(𝑑𝑦𝑑𝑥

)

[2 + (𝑑𝑧𝑑𝑥

)2

]

32

𝑑𝑧𝑑𝑥

= 8

𝜕2𝑦

𝜕𝑢2+ 6

𝜕2𝑦

𝜕𝑥2= 0

𝑤ℎ𝑒𝑟𝑒: 𝑦 = 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒; 𝑢 & 𝑥 = 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠

𝑥𝜕𝑢

𝜕𝑥+ 6𝑦

𝜕𝑦

𝜕𝑥= 𝑛𝑢

Types

a) Ordinary differential equation

b) Partial differential equation

ODE: A differential equation is an ordinary differential equation if the unknown function

depends on only one independent variable. That is, if the derivatives in it are with respect to a

single independent variable



Examples 𝑑𝑦

𝑑𝑥= 3𝑥 + 8

𝑑4𝑦

𝑑𝑢4+

𝑑2𝑦

𝑑𝑢2+ 6

𝑑𝑦

𝑑𝑢+ 𝑎𝑦 = 0

𝑑3𝑦

𝑑𝑝3+ (

𝑑𝑦

𝑑𝑝)

4

+ 3𝑝𝑦 = 7

𝑒𝑦𝑑2𝑦

𝑑𝑥2+ 5 (

𝑑𝑦

𝑑𝑥)

2

= 1

8𝑑3𝑦

𝑑𝑥3+ (𝑐𝑜𝑠𝑥)

𝑑2𝑦

𝑑𝑥2+ 9𝑥𝑦 = 0

PDE: If the unknown function depends on two or more independent variables, the differential

equation is a partial differential equation. That is, if the derivatives in it are with respect to more

than one independent variable

Examples

𝜕2𝑦

𝜕𝑡2+ 6

𝜕2𝑦

𝜕𝑥2= 0

𝜕2𝑣

𝜕𝑥2+

𝜕2𝑣

𝜕𝑢2+ 4

𝜕𝑣

𝜕𝑡= 7

Order

This is order of the highest derivative in the differential equation.

Degree



The degree of a differential equation is power of the highest order derivative term in the differential

equation.

Examples

𝑑2𝑧

𝑑𝑥2+ 5𝑥 (

𝑑𝑧

𝑑𝑥)

4

+ 3𝑧 = 𝑠𝑒𝑐𝑥, 𝑑𝑒𝑔𝑟𝑒𝑒 = 1

𝑑𝑦

𝑑𝑥= 𝑥 +

𝑘

(𝑑𝑦𝑑𝑥

) 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦

∴ (𝑑𝑦

𝑑𝑥)

2

= 𝑥𝑑𝑦

𝑑𝑥+ 𝑘 𝑑𝑒𝑔𝑟𝑒𝑒 = 2

[4 + (𝑑𝑧𝑑𝑣

)2

]

34

𝑑2𝑧𝑑2𝑣

= 2

[4 + (𝑑𝑧

𝑑𝑣)

2

]

3

= (2𝑑2𝑧

𝑑2𝑣)

4

𝑑𝑒𝑔𝑟𝑒𝑒 = 4

Linearity

A differential equation is linear, if

Dependent variable and its derivatives are of degree one,

Coefficients of term do not depend upon dependent variable.

Forming differential equation

Eliminating arbitrary constants from the given function

1. xBxAy cossin

xBxAdx

dysincos



xBxAdx

ydcossin

2

2

)cossin(2

2

xBxAdx

yd

ydx

yd

2

2

02

2

ydx

yd

2. 2xBeAexy xx

xBeAedx

dyxy xx 2

22

2

xx BeAedx

ydx

dx

dy

dx

dy

But, 2xxyBeAe xx

Thus, 22 2

2

2

xxydx

ydx

dx

dy

In general an nth order differential equation will result from consideration of a function with n arbitrary

constants.

Solution of differential equation

Direct integration

Separating the variables

Homogeneous equations – by substituting y = vx

Linear equations – use of integrating factor

Direct integration



dxxfydxxfdydxxfdyfxfdx

dy)()()()(

Examples

3. 563 2 xx

dx

dy

dxxxdy )563( 2

dxxxdy )563( 2

cxxxy 53 23

Separating the variable

Examples

2 2

2

2

2

1

( 1) 2 ( 1) 2

2

22 2

2 2 2 (Let 2 )

2 2

dy x

dx y

y dy xdx y dy xdx

ydy dy xdx

y xy C

y y x C C A

y y x A

Homogenous differential equations

The equation whose total degree in X and Y for the terms involved is one

The substitution of ( ) is used or ( ).y

v x y v x xx

4.

Example

3

2

dy x y

dx x



(1 3

2

yx

dy x

dx x

(1 3

.......2

y

dy xi

dx

Let ( )y

V y V x Xx

.......dy dv

v x iidx dx

substitute (ii) into (i)

1 3

2

dv vv x

dx

1 3

2

dv vx v

dx

1 3 2

2

dv v vx

dx

1 (separate the variables)

2

dv vx

dx

2 (Integrate)

1

dxdv

v x

1 12

1dv dx

v x

2ln 1 ln lnv x c

2

1 since and ln ln ln( )y y

cx v x c xcxx

Tautolia

5. 2

2

1

1

x

y

dx

dy

separate the variables



22 11 x

dx

y

dy

Integrate

22 11 x

dx

y

dyFrom xdx

x

1

2sin

1

1

cxy 11 sinsin

6. ydyyxxdxyx )823()732( 2222

ydyyxxdxyx )823()732( 2222

)732(

)823(22

22

yx

yx

ydy

xdx

From componendo rule d

dc

b

ba

c

c

b

a

Dividendo rule d

dc

b

ba

c

c

b

a

Componendo and dividendo rule dc

dc

ba

ba

Thus

ydyxdx

ydyxdx

)732()823(

)732()823(2222

2222

yxyx

yxyx

1

)3(522

22

yx

yx

ydyxdx

ydyxdx

1

)(5

3 2222

yx

ydyxdx

yx

ydyxdx

Multiply by 2 both sides to the numerators

1

)22(5

3

222222

yx

ydyxdx

yx

ydyxdx Integrate

1

)22(5

3

222222 yx

ydyxdx

yx

ydyxdx



cyxyx ln)1ln(5)3ln( 2222

52222 )1()3( yxcyx

7. yyx exedx

dy 2

22 1

xeee

x

e

e

dx

dy x

yyy

x

Seperate the variables

dxxedye xy 2 Integrate

dxxedye xy 2

cxee xy 3

3

1

8. 22a

dx

dyyx

Let dx

dy

dx

dvyxv 1 …….ii

1dx

dv

dx

dy

22 1 adx

dvv

2

22

2

2

1v

va

dx

dv

v

a

dx

dv Separate the variable

dxdyva

v

22

2

By dividing 222 )( vva

dxdvva

a

22

2

1 Integrate

dxdv

va

a22

2

1



dxdv

a

vdv

2

1

1

cxa

vav 1tan But yxv

cxa

yxayx

1tan

cxa

yxay

1tan

9. )cos()sin( yxyxdx

dy

Let dx

dy

dx

dvyxv 1

1dx

dv

dx

dy

vvdx

dvcossin1

1cossin vvdx

dv Separate the variables

dxvv

dv

1cossin

But: 2

cos2

sin2sin vvv and 12

cos2cos 2 vv

dxvvv

dv

112

cos22

cos2

sin2 2

dx

v

vv

dv

1

2cos

2sin

2cos2 2



dx

v

dvv

1

2tan2

2sec2

Integrate

dxv

dvv

12

tan

2sec

2

12

If we let

2sec

21

2tan

2 v

dwdvvw

cxv 12

tanln But yxv

cxyx

2

1ln

10.

byx

ayx

dx

dy

byx

ayx

dx

dy

dx

dvyxv 1

1dx

dv

dx

dy Substitute

bv

av

dx

dv

bv

av1 Divide by

bv

av

av

bv

bv

av

dx

dv1

abbvavv

abavvbv

dx

dv2

2

1

12

2

abbvavv

abavvbv

dx

dv

abbvavv

abbvavvabavvbv

dx

dv

2

22

avbvabv

abv

dx

dv

2

2 )(2 Separate the variables



dxdvabv

vababv2

)(2

2

dxdvabv

vab

abv

abv2

)(22

2

dxdvabv

vab2

)(1

2

Integrate

dxdvabv

vabdv 2)(

2

cxabvabv 2ln2

1)( 2 But yxv

cxabyxabyx 2)(ln2

1)( 2

cxabyxab

y

2)(ln2

)(

11. 214 yxdx

dy

Let 14 yxv

4dx

dv

dx

dy

dx

dy Substitute

24 vdx

dv

24 vdx


dxv

dv

24 Integrate

dxv

dv2

214



cxv

2tanln

2

1 1 But 14 yxv

cxyx

2

14tanln

2

1 1

12. dyyyxdxxxy )tanlog(seccos)tanlog(seccos

Divide by yxcoscos both sides to get

yx

yyxdx

yx

xxy

coscos

)tanlog(seccos

coscos

)tanlog(seccos

dyyyydxxxx )tanlog(secsec)tanlog(secsec

dyyyydxxxx )tanlog(secsec)tanlog(secsec

By letting )tanlog(sec xxw and )tanlog(sec yyz

dxx

dwx

dx

dw

secsec

dyy

dzx

dy

dz

secsec

dzy

zydw

x

wx

sec

)(sec

sec

)(sec

czwczw 2222

2

1

2

1 Substitute values of zw,

cyyxx 22 )tanlog(sec)tanlog(sec

The linear d.e is in the form of

Qpydx

dy Or Qpx

dy

dx

Where the Integrating factor is pdx

eIF or pdy

eIF



General solution CIFdxQIFY ..

13. xeydx

dy 25

xeQp 25

xdxpdx

eeeFI 55

.

From CIFdxQIFY ..

Cdxeeye xxx 525

Cdxeye xx 75

ceye xx 75

7

1

xx ceey 52

7

1

14. 02

2

dxdyxye

y

Find the value of so that 2ye is an integrating factor of the given

differential equation

22

22 yy

exydy

dx

dy

dxxye

From Qpxdy

dx ; yp , 2

2y

eQ

and pdy

eFI.

2

2

.

yydy

eeFI

Compare the two integrating factors 2

2

2y

y ee

2

1

2

22

yy



15. Given. Qypydx

dy Find the general solution

Since pdx

eFI. Multiply the pdx

eFI. both sides

pdxpdxpdx

Qepyedx

dye

dxQeyedQedx

yedpdxpdxpdx

pdx

.

.

pdx

pdx

pdxpdx

e

cdxQeydxQecye .

AdxQeeypdxpdx

where cA

Bernoulli’s equation

16. nQypy

dx

dy

We need to make it linear

nQypydx

dy Divide by ny both sides

n

n

nn y

Qy

y

py

y

dx

dy

Qpydx

dyy nb 1

Let nyz 1

dx

dyyn

dx

dz n )1( Divide by )1( n

dx

dz

ndx

dyy n

)1(

1

Substitute



Qpzdx

dz

n

)1(

1 Multiply by )1( n

Qnpzndx

dz)1()1(

Let pnp )1(1 and QnQ )1(1 Substitute

11 Qzpdx

dz Made linear

17. Find the general solution of 11 Qzpdx

dz

Where pnp )1(1 ; QnQ )1(1 and nyz 1

pdxndxp

eeFI)1(1

.

From. CIFdxQIFZCIFdxQIFY .... 1

cdxeQeZpdxnpdxn

)1(

1

)1(

.

cdxeQneypdxnpdxnn

)1()1(1 .)1(. Divide by pdxn

e)1(

both sides

cdxeQneypdxnpdxnn )1()1()1(1 .)1( Multiply by

n1

1both sides

npdxnpdxn

cdxeQney

1

1

)1()1(

.)1(

18. xyxdx

dyy 2coscostantan

y

xy

y

x

dx

dz

y

y

cos

coscos

cos

tan

cos

tan 2

xyxdx

dzyy 2cossectantansec

Let yz sec



dx

dyyy

dx

dztansec Substitute

xzxdx

dz 2cos.tan Compare

Qpydx

dy : xQxp 2costan

xeeeeFI xxxxdx

sec. seclncos

1ln

coslntan

From CIFdxQIFZ ..

Cxxxz seccossec 2

Cxx

zx

2coscos

1

cos

1

cxCxzx

sincoscos

1

)(sincos cxxz But yz sec

)(sincossec cxxy

19. dxyxdyyx )123()146(

1)23(2

123

146

123

yx

yx

yx

yx

dx

dy

Let yxv 23

dx

dv

dx

dy

dx

dy

dx

dv3

2

123 Substitute

12

13

2

1

v

v

dx

dv

)12(

)22(3

v

v

dx

dv



)12(

)22(3

v

v

dx

dv

12

14

)12(

)2236(

v

v

v

vv

dx


dxdvv

v

14

12 Divide 12)14( vv

dxdvv

)14(2

1

2

1 Integrate

dxdvv

dv)14(

1

2

1

2

1

cxvv )14ln(4

1.

2

1

2

1 But yxv 23

cxyxyx

)1)23(4ln(8

1

2

23

cxyxyx

)1812ln(8

1

2

23

20. yydx

dyx

1

x

yy

dx

dy

1

yx

y

dx

dy

1

x

xyy

dx

dy

1

x

xy

dx

dy )1(1

x

yx

xdx

dy )1(1



xx

yx

dx

dy 1)1(

xy

xdx

dy 11

1

Compare with

Qpydx

dy :

xQ

xp

11

1

xxxxxdxdx

xdx

xpdx

xeeeeeeeFI

.. lnln

11

1

From CIFdxQIFy ..

cecdxxex

yxe xxx 1

xxe

c

xy

1

21. Set of function x

x1

, form a basis of solution

02 yyxyx Condition ;1)1( y 2)1( y

Solution

xBAxy

1 Differentiate

2

1

xBAy ……………….i

3

12

xBy ……………………ii

iiiyx

B .................2

3

2

1.

2 2

3 yxA

x

yxAy

ivyx

yA .......2



x

yxx

yxyy

1

22

3

2

2

22

2 2222 yxyxyxyxyxyxy

yxyxyxyxyxyx

y

2

22

22

22

2

22

02 yyxyx Shown

Solve the values of A and B given that ;1)1( y 2)1( y

xBAxy

1 vBA ................1

2

1

xBAy viBA .................2

By solving simultaneous equations: 2

3A and

2

1B

xxy

2

1

2

3 Particular solution

22. Given that Qpydx

dy To show that

pdx

eFI.

Solution

Let )(xvv be the Integrating Factor FI.

iQxvpyxvdx

dyxv ...........)()()(

If the L.H.S of the equation is exact D.E, then

iiQxvdx

dvy

dx

dyxv

dx

yvd.........)(.)(

),(

Compare two equations



vpdx

dvvpy

dx

dvy

pdxdvv

pdxv

dv 1

pdxvpdxv elogln

pdx

eFIxV .)(

23. xxxyxdx

dyxx 3523 13 Solve the differential equation

xx

xxxy

xx

x

dx

dy

3

35

3

2 13

xx

xP

3

2 13 And

xx

xxxQ

3

35

xxeeeeFI xxxx

dxxx

xpdx

3

1ln

)ln(

131

.333

2

From CIFdxQIFY ..

cdxxxxx

xxx

xxy

33

35

3

1.

1.

cdx

xxxx

xx

xxy

)1(

1.

)1(

)1(1.

22

22

3

cxcdxx

cdxxx

xx

xxy

ln1

)1(

)1(1.

222

22

3

cxxxy ln3

EQUATIONS REDUCIBLE TO HOMOGENEOUS FORM



Case 1: CByAx

cbyax

dx

dy

If

B

b

A

a Just use normal substitution

24. 146

123

yx

yx

dx

dy

1)23(2

123

yx

yx

dx

dy 6;3 Aa ; 4;2 Bb And

2

1

B

b

A

a

Let

dx

dv

dx

dy

dx

dy

dx

dvyxv 3

2

12323

12

223

1)(2

13

2

1

v

v

dx

dv

v

v

dx

dv

12

2236

12

223

v

vv

v

v

dx

dv

dxdvv

v

v

v

dx

dv

14

12

12

14 But

)14(2

1

2

1

14

2

1

2

12

1

2

1

2

12

12)14(14

12

vv

v

vvv

v

dxdv

v )14(2

1

2

1

dxdvv

dv14

1

2

1

2

1

cxvv 14ln8

1

2

1 But yxv 23

cxvv 14ln8

1

2

1

25. dydxdydxyx ))(2(

dydxydyydxxdyxdx 22

dyydyxdydxydxxdx 22

)12()12( yxdyyxdx



12

12

yx

yx

dx

dy 1;1 Aa ; 2;2 Bb And 1

B

b

A

a

Let

1

2

1212

dx

dv

dx

dy

dx

dy

dx

dvyxv

1

221

1

11

2

1

v

v

dx

dv

v

v

dx

dv

1

131

1

22

v

v

v

v

dx

dv

dxdvv

v

13

1 But

)13(3

4

3

1

13

1

vv

v

cxvvdxdvv

13ln

9

4

3

1

)13(3

4

3

1 But yxv 2

cxyxyx

163ln9

4

3

2


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