Higher Engineering Mathematics

Ess-For-H8152.tex 19/7/2006 18: 2 Page 716

716 ESSENTIAL FORMULAE

Numerical solutions of first order differential

equations

Euler’s method: y 1 =y 0 +h(y′) 0

Euler-Cauchy method: yP 1 =y 0 +h(y′) 0

and yC 1 =y 0 +

1

2

h[(y′) 0 +f(x 1 ,yp 1 )]

Runge-Kutta method:

To solve the differential equation

dy

dx

=f(x,y)given

the initial conditiony=y 0 atx=x 0 for a range of

values ofx=x 0 (h)xn:

1. Identifyx 0 ,y 0 andh, and values ofx 1 ,x 2 ,x 3 ,...


2. Evaluatek 1 =f(xn,yn) starting withn= 0


3. Evaluatek 2 =f

(

xn+

h

2

,yn+

h

2

k 1

)

4. Evaluatek 3 =f

(

xn+

h

2

,yn+

h

2

k 2

)

5. Evaluatek 4 =f(xn+h,yn+hk 3 )


6. Use the values determined from steps 2 to 5 to
   evaluate:

yn+ 1 =yn+

h

6

{k 1 + 2 k 2 + 2 k 3 +k 4 }

7. Repeat steps 2 to 6 forn=1, 2, 3,...

Second order differential equations

Ifa

d^2 y

dx^2

+b

dy

dx

+cy= 0 (wherea,bandcare

constants) then:

(i) rewrite the differential equation as

(aD^2 +bD+c)y= 0

(ii) substitute mfor D and solve the auxiliary

equationam^2 +bm+c= 0

(iii) if the roots of the auxiliary equation are:

(a)real and different, saym=αandm=β

then the general solution is

y=Aeαx+Beβx

(b)real and equal, saym=αtwice, then the

general solution is

y=(Ax+B)eαx

(c)complex, saym=α±jβ, then the general

solution is

y=eαx(Acosβx+Bsinβx)

(iv) given boundary conditions, constantsAandB

can be determined and the particular solution

obtained.

Ifa

d^2 y

dx^2

+b

dy

dx

+cy=f(x)then:

(i) rewrite the differential equation as

(aD^2 +bD+c)y=0.

(ii) substitute mfor D and solve the auxiliary

equationam^2 +bm+c=0.

(iii) obtain the complimentary function (C.F.),u,as

per (iii) above.

(iv) to find the particular integral,v, first assume a

particular integral which is suggested byf(x),

but which contains undetermined coefficients

(See Table 51.1, page 482 for guidance).

(v) substitute the suggested particular integral into

the original differential equation and equate

relevant coefficients to find the constants

introduced.

(vi) the general solution is given byy=u+v.

(vii) given boundary conditions, arbitrary constants

in the C.F. can be determined and the particular

solution obtained.

Higher derivatives

y y(n)

eax aneax

sinax ansin

(

ax+

nπ

2

)

cosax ancos

(

ax+

nπ

2

)

xa

a!

(a−n)!

xa−n

sinhax

an

2

{[ 1 +(−1)n]sinhax

+[ 1 −(−1)n]coshax}

coshax

an

2

{[1−(−1)n] sinhax

+[1+(−1)n] coshax}

lnax (−1)n−^1

(n−1)!

xn