Higher Engineering Mathematics, Sixth Edition

670 Higher Engineering Mathematics

(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) substitutemforDandsolvetheauxiliaryequation

am^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 par-

ticular integral which is suggested byf(x),but

which contains undetermined coefficients (See

Table 51.1, page 484 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

)

y y(n)

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

Leibniz’s theorem:

To find then’th derivative of a producty=uv:

y(n)=(uv)(n)=u(n)v+nu(n−^1 )v(^1 )

+

n(n− 1 )

2!

u(n−^2 )v(^2 )

+

n(n− 1 )(n− 2 )

3!

u(n−^3 )v(^3 )+···

Power series solutions of second order

differential equations:

(a) Leibniz-Maclaurin method

(i) Differentiate the given equationntimes,

using the Leibniz theorem,

(ii) rearrange the result to obtain the recurrence

relation atx=0,

(iii) determine the values of the derivatives at

x=0, i.e. find(y) 0 and(y′) 0 ,

(iv) substitute in the Maclaurin expansion for

y=f(x),

(v) simplifythe result where possible and apply

boundary condition (if given).

(b) Frobenius method

(i) Assume a trial solution of the form:

y=xc{a 0 +a 1 x+a 2 x^2 +a 3 x^3 +···+

arxr+···} a 0 =0,