Linear partial differential equations of second and higher order with constant coefficients of both homogeneous and non-homogeneous types

LINEAR PARTIAL DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER WITH CONSTANT COEFFICIENTS OF BOTH HOMOGENEOUS AND NON-HOMOGENEOUS TYPES

1. Partial differential equation of Higher order

Linear partial differential equations of higher order with constant co-efficients may be divided into two categories as given below. 

(i) homogeneous p.d.e. with constant co-efficients. 

(ii) non-homogeneous p.d.e. with constant co-efficients.

Definition :

A linear p.d.e. with constant co-efficients in which all the partial derivatives are of the same order is called homogeneous; otherwise it is called non-homogeneous.

2. Homogeneous linear equation

A homogeneous linear partial differential equation of nth order with constant co-efficients is of the form

equation (i) can be written as

Method of finding C.F.

1. To get the auxiliary equation of f(D, D') = F(x, y)

put D = m and D' = 1.

2. The auxiliary equation is f(D,D') = 0

i.e., f(m, 1) = 0

To find the complementary functions:

Note: To find :

Problems based on Homogeneous equations 

Example 1.5.1: Solve (D2 – 4DD' + 3D'2) z = 0. 

Solution: 

Given [D2 - 4DD' + 3D'2] z = 0

The auxiliary equation is m2 - 4m +3 = 0

[Replace D by m and D' by 1]

Solving, m2 - 3m - m + 3

m (m - 3) - 1 (m − 3) = 0

(m − 1) (m − 3) = 0

m = 1, m = 3

Here, the roots are distinct

Since, R.H.S. is zero, there is no particular integral

Hence, the general solution is z = C.F.

Example 1.5.2: Solve [D2 - 2DD' + D'2] z = 0.

Solution : [D2 - 2DD' + D'2]z = 0.

The auxiliary equation is m2 − 2m + 1 = 0

[Replace D by m and D' by 1]

i.e., (m − 1)2 = 0

m = 1, 1

Here, the roots are equal

Since R.H.S. is zero, there is no particular integral

Hence, the general solution is z = C.F

Example 1.5.3: Solve [D3 + DD′2 – D2 D′ – D'3] z = 0

Solution: Given [D3 + DD'2 – D2 D' - D'3]z = 0

The auxiliary equation is m3 - m2 + m −1 = 0

[Replace D by m and D' by 1]

Here, the roots are distinct

Since R.H.S. is zero, there is no particular integral

Hence, the general solution is

Example 1.5.4: Solve 2r+ 5s - 3t = 0.

Solution: The given differential equation can be written as

The auxiliary equation is 2m2 + 5m - 3 = 0

[Replace D by m and D' by 1]

2m2 + 6m - m −3 = 0

2m (m + 3) - 1 (m + 3) = 0

(m + 3) (2m − 1) = 0

Here the roots are distinct.

Since R.H.S. is zero, there is no particular integral

Hence, the general solution is

Example 1.5.5 : Solve (D4 – D'4) z = 0

Solution: 

Here, the roots are distinct

Since R.H.S. is zero, there is no particular integral

⸫ Hence, the general solution is

EXERCISES

I. Solve : [R.H.S = 0, unequal roots]

II. Solve [R.H.S = 0, unequal roots of Higher order].

III. Solve : [R.H.S = 0, equal roots]

IV. Solve : [R.H.S= 0, different types of roots]

Note: To solve [DD' – D'2]z = 0

R.H.S eax+by Replace D by a and D' by b

Example 1.5.6 : Solve :

Solution: 

Example 1.5.7: Find the P.I of [D2 + 4DD'] z = ex.

Solution: 

Example 1.5.8 : Solve

Solution: 

Example 1.5.9: Solve (D + D')2 z = ex-y

Solution: Given: (D+D')2z = ex-y

The auxiliary equation is

Example 1.5.10: Find the P.I of (D2 + DD') z = ex-y

Solution: 

Example 1.5.11: Solve (D3 – 3DD'2 + 2D'3) z = e2x − y + ex+y

Solution: 

Hence, the general solution is

EXERCISES

Find the P.I of the following :

R.H.S = x2 y2

Example 1.5.12: Solve [D2 - 7DD' + 6D'2] z = xy

Solution :

Given: [D2 - 7 DD' + 6D'2]z = xy

The auxiliary equation is

Example 1.5.13 : Solve [D2 + 3 DD' + 2D'2] z = x + y

Solution: Given: [D2 + 3 DD' + 2D'2] z = x + y

The auxiliary equation is

Example 1.5.14: Solve [D2 + 4DD' - 5D'2] z = x + y2 + π

Solution: Given: [D2 + 4DD' - 5D'2]z = x + у2 + π

The auxiliary equation is

EXERCISES

Find the Particular Integral of 

R.H.S = eax+by + xrys

Example 1.5.15 : Solve: [D2 - DD' – 6D'2] z = x2y + e3x + y.

Solution: 

Example 1.5.16: Solve (D2 + DD' - 6D'2) z = x2y + e3x+y.

Solution :

Example 1.5.17 : Solve (D2 – 6DD' + 5D'2 ) z = ex sinh y + xy

Solution: 

Example 1.5.18: Solve (D2 + 2DD' + D'2) z = x2y + ex-y

Solution :

EXERCISES

Find the Particular Integral of the following :

R.H.S = sin (ax + by) or cos (ax + by)

Replace D2 by - a2, DD' by - ab, D'2 by - b2

Formulae

Example 1.5.19 : Solve [D2 - 2DD' + 2D'2] z = sin (x − y).

Solution: 

Example 1.5.20: Solve [D3 - 4D2D' + 4DD'2] z = 6 sin (3x + 6y). 

Solution :

Given: [D3 - 4D2 D' + 4DD'2] z = 6 sin (3x + 6y)

The auxiliary equation is m3 - 4m2 + 4i = 0

Example 1.5.21 : Solve [D2 – 2DD' + D'2] z = cos (x − 3y). 

Solution: Given: [D2 − 2DD' + D'2]z = cos (x - 3y)

Example 1.5 22 : Solve

Solution: 

Example 1.5.23: Solve [D2 – 3DD' + 2D'2] z = sin x cos y

Solution :

Given: [D2 - 3DD' + 2D'2] z = sin x cos y

The auxiliary equation is m2 - 3m + 2 = 0

m = 1, 2

Example 1.5.24: Solve (D2 - DD') z = sin x sin 2y

Solution: Given: (D2 - DD')z = sinx sin 2y

The auxiliary equation is m2 - m = 0

EXERCISES

Find the P.I of the following:

R.H.S= sin (ax + by) + eax+by

Example 1.5.25: Solve (D3 - 7DD'2 – 6D'3) z = sin(x+2y) + e3x+y.

Solution: 

Example 1.5.26 : Solve :

Sol. 

EXERCISES

Find the P.I. of the following:

R.H.S = xr ys + sin (ax + by) or cos (ax + by)

Example 1.5.27: Solve (D2 + 3DD' – 4 (D')2) z = x + sin y 

Solution: 

Example 1.5.28: Solve the equation

[D3 – 7 DD'2 – 6 D'3] z = cos (x + 2y) + x

Solution: 

R.H.S = eax+by ϕ(x, y)

Replace D by D+a, D' by D'+b

Example 1.5.29 : Solve [D2 − 2DD' + D'2] z = x2 y2 ex + y. 

Solution : 

Example 1.5.30 : Solve (D3 + D2 D' - DD'2 – D'3) z = excos 2y.

Solution: 

The general solution is z = C.F + P.I

EXERCISES

R.H.S. = sin ax sin by (or) cos ax cos by

Example 1.5.31: Solve [D2 - D'2] z = sin 2x sin 3y 

Solution: 

R.H.S = y cos.x (or) y sin x

Example 1.5.32 : Solve [D2 + DD' - 6D'2] z = y cos x. (or) (r+ s − 6t) = y cos x

Solution: 

Aliter: To find P.I

EXERCISES

3. Non-homogeneous linear equation

The linear differential equations which are not homogeneous, are called Non-homogeneous linear equations.

Complementary function :

Let the non-homogeneous equation be (D – mD' — a) z = 0

Similarly, the solution of (D - mD' - a)2 z = 0 is

Working rule

Problems based on Non-homogeneous linear equation

Example 1.5.34: Solve (D + D' - 2) z = 0

Solution : 

Given : (D + D' − 2) z = 0

i.e., (D − (−1) D' − 2] z = 0

We know that, working rule case (i) is

If (D - m D' - c) z = 0, then z = ecx f(y+mx)

where ƒ is arbitrary.

Example 1.5.35 : Solve (D + D′ − 2) (D + 4 D' − 3 ) z = 0 

Solution: Given: (D + D' - 2) (D + 4D' − 3) z = 0

i.e., [D − (−1) D' − 2 ] [ D − (−4) D' — 3] z = 0

We know that, working rule case (ii) is

Example 1.5.36 : Solve : (D2 – DD' + D' − 1) z′ = 0

Solution :

Another method :

Example 1.5.37 : Solve : (2 DD' + D'2 – 3D′) z = 3 cos (3x — 2y)

Solution :

To find the complementary function

Another method to find C.F.

Example 1.5.38 : Solve (D − D' − 1) (D − D' − 2) z = e2x−y

Solution :

To find C.F. take

 (D − D' − 1) (D − D' − 2) z = 0

by working rule case (ii)

Example 1.5.39: Solve (D − D' − 1) (D − D' — 2) z = e2x+y

Solution: Given: (D − D' − 1) (D − D' - 2) z = e2x+y

To find C.F.

 (D − D' − 1) (D − D' – 2) z = 0

by working rule case (ii)

Example 1.5.40 : Solve (D + D' − 1) (D + 2D' − 3) z = 4 + 3x + 6y.

Solution :

Given (D + D' − 1) (D + 2D' - 3) z = 4 + 3x + 6y

To find C.F.

(D + D' − 1) (D + 2D' - 3) z = 0

(D − (−1) D' − 1) (D − (−2) D' - 3) z = 0

by working rule case (ii)

Example 1.5.41: Solve (D + 3D' + 4)2 z = 0

Solution: Given: (D + 3D' + 4)2 z = 0

 (D − (−3) D' − (−4)]2 z = 0

by working rule case (iii)

Example 1.5.42 : Find the general solution of

(D2 - 3DD' + 2D'2 + 2D − 2D') z = sin (2x + y) 

Solution: Given: (D2 – 3DD' + 2D'2 + 2D - 2D') z = sin (2x+y)

EXERCISES

Answers: