Lagrange's Linear Equation

LAGRANGE'S LINEAR EQUATION

The equation of the form Pp + Q q = R = R is known as Lagrange's equation when P, Q & R are functions of x, y and z.

To solve this equation it is enough to solve the subsidiary equation

Working Rule

First step. Write down the subsidiary equations

Second step: Solve the above subsidiary equations.

Let the two solutions be u = a and v = b

Third step Then f(u, v) = 0 or u = ϕ(v) is the required solution of P p + Qq = R

Generally, the subsidiary equation can be solved in two ways.

1. Method of Grouping

2. Method of Multipliers

1. Method of Grouping

In the subsidiary equation if the variables can be separated in any pair of equations, then we get a solution of the form u (x, y) = a and v (x, y) = b.

2. Method of Multipliers

Choose any three multipliers l, m, n which may be constants or function of x, y, z we have

If it is possible to choose 1, m, n such that lP + mQ+ nR = 0

then ldx + mdy + ndz = 0

If ldx + mdy + ndz is an exact differential then on integration we get a solution u = a.

The multipliers l, m, n are called lagrangian multipliers.

1.4a. Problems based on Lagrange's linear equation method of Grouping :

Example 1.4a(1): Solve px + qy = z.

Solution: 

Given: px + qy = z (i.e.,) xp + yq= z …..(1)

This equation is of the form Pp + Qq = R

where P = x, Q = y, R = z

Example 1.4a(2): Write the solution of px2 + qy2 = z2.

Solution: 

Given: px2 + qy2 = z2

i.e., x2p + y2q = z2

This equation is of the form P p + Q q = R

where P = x2, Q = y2, R = z2

Example 1.4a(3) Find the solution of p√x + q√y = √z.

Solution: 

Example 1.4a(4) : Find the general solution of p tan x + q tan y = tan z.

Solution :

Example 1.4a(5): Write the general integral of pyz + qzx = xy.

Solution: 

Given: pyz + qzx = xy (i.e.,) yzp + zxq = xy

This equation is of the form Pp + Qq = R

where P = yz, Q = zx, R = xy

The Lagrange's subsidiary equations are

Hence, the general solution is ƒ (u, v) = 0

i.e., f (x2 - y2, y2 - z2) = 0, where ƒ is arbitrary.

Example 1.4a(6) : Find the general integral of p − q = log (x + y).

Solution :

Example 1.4a (7): Obtain the general solution of pzx+qzy = xy.

Solution: 

Example 1.4a(8) : Solve :y2 p - xyq = x (z − 2y)

Solution: 

Example 1.4.a(9): Solve x2p + y2 q = z

Solution: 

EXERCISES 1.4.a

Solve the following :

1.4.b. Problems based on Lagrange's method of multipliers

Example 1.4b(1): Solve x (y-z) p + y (z - x) q=z (x − y).

(OR) Solve

Solution: 

Given: x (y − z) p + y (z − x) q = z (x − y)

This equation is of the form P p + Q q = R

Taking the Lagrangian multipliers are 1, 1, 1, we get each ratio in (1)

Hence, d(x + y + z) = 0 Integrating, we get

x + y + z = a

Taking the Lagrangian multipliers 1/x, 1/y, 1/z, we get each ratio in (1)

Integrating, we get

logx + logy + log z = log b

log (xyz) = log b

xyz = b

Hence, the general solution is f (a, b) = 0

i.e., f (x + y + z, xyz) = 0

where f is arbitrary.

Example 1.4b(2): Solve (mz - ny) p + (nx - lz) q = ly - mx. 

Solution: 

Given : (mz - ny) p + (nx - lz) q = ly - mx

This equation is of the form Pp + Qq = R

Taking the Lagrangian multipliers are x, y, z, we get each ratio in (1)

Hence, xdx + ydy + zdz = 0

Integrating, we get

Taking the Lagrangian multipliers are 1, m, n, we get each ratio in (1)

Hence, ldx + mdy + ndz = 0

Integrating, we get

lx + my + nz =b

lx + my + nz = b

Hence, the general solution is ƒ (a, b) = 0

i.e., f(x2 + y2 + z2, lx + my + nz) = 0, where ƒ is arbitrary.

Example 1.4b(3): Solve (3z - 4y) p + (4x-2z) q 4y) p + (4x - 2z) q = 2y - 3x.

Solution:

Use Lagrangian multipliers x, y, z,

we get each ratio in (1)

Again use Lagrangian multipliers 2, 3, 4,

we get each ratio in (1)

Hence, the general solution is f (a, b) = 0

i.e., f(x2 + y2 + z2, 2x + 3y + 4z) = 0, where ƒ is arbitrary.

Example 1.4b(4) Find the general solution of x2 (y2 – z2 ) p + y (z2 − x2) q = z (x2 − y2) (or) x (z2 − y2) p + y (x2 − z2) q = z (y2 − x2)

Solution: 

Use Lagrange multipliers x, y, z, we get each ratio in (2)

logx + logy + log z = log b

log (xyz) = log b

i.e., xyz = b

Hence, the general solution is ƒ (a, b) = 0 i.e., f (x2 + y2 + z2, xyz) = 0,

where f is arbitrary.

Example 1.4b(5): Solve (x2 - yz) p + (y2 − zx) q = (z2 − xy).

Solution: 

a = x2 + y2 + z2 − (x + y + z)2

= x2 + y2 + z2 - x2 - y2 - z2 - 2xy - 2yz - 2zx

= -2 (xy + yz + zx)

Using two sets of multipliers 1, −1, 0; 0, 1, −1 each of the ratio in (2).

Integrating on both sides, we get

log (x − y) = log (y - z) + log b

log (x − y) = log [b (y - z)]

x – y = b (y - z)

Aliter: Using Lagrange multipliers y + z, z + x, x + y we get each ratio in (2)

Note: The author wishes to thank Dr. B. Jothiram, formerly Asst. Prof. and HOD Maths, Govt. College of Engg., Salem, for having drawn his attention to this elegant method of getting one of the independent solutions.

Example 1.4b(6): Solve (x2 + y2 + yz) p + (x2 + y2 − xz) q = z(x + y). 

Solution : 

Use Lagrange multipliers 1, -1, -1, we get each ratio in (1)

Integrating, we get

Integrating, we get

Example 1.4b(7): Solve (y2 + z2) p – xyq + xz = 0

Solution: 

Use Lagrangian multipliers x, y, z,

we get each ratio in (A)

Example 1.4b(8): Solve : x (y2 + z) p + y (x2 + z) q = z (x2 - y2).

Solution: This equation is of the form P p + Q q = R

where P = x (y2 + z), Q = y (x2 + z), R = z (x2 - y2).

The Lagrange's subsidiary equations are

Taking the Lagrangian multipliers x, −y, −1, we get

Hence, x dx - y dy - dz = 0

Integrating, we get

Example 1.4b(9): Solve the equation: (x2 – y2 – z2) p + 2 xyq = 2 zx

Solution: 

Choosing the method of multiplier idea x, y, z as multipliers.

Example 1.4b(10): Solve (z2 - 2yz - y2) p + (xy + zx) q = xy - zx

Solution: 

The Lagrange's subsidiary equations are

Choose the multipliers x, y, z, we get.

Hence, the general solution is

ƒ (y2 - 2yz - z2, x2 + y2 + z2) = 0, where ƒ is arbitrary.

Example 1.4b(11) : Solve (y + z) p + (z + x) q = x + y.

Solution :

Given (y + z) p + (z + x) q = x + y

This equation is of the form Pp + Qq = R

Hence, the general solution is ƒ (u, v) = 0

Example 1.4b(12): Solve (y-xz) p + (yz - x) q = (x + y) (x − y)

Sol.

Choose Lagrangian multipliers x, y, z,

Choose Lagrangian multipliers y, x, 1,

Example 1.4b(13) : Solve (y − z) p − (2x + y) q = 2x + z.

Solution: Given: (y - z)p (2x + y) q = 2x + z

This equation is of the form Pp + Qq = R

where P = y – z, Q = -(2x + y), R = 2x + z

The Lagrange's subsidiary equations are,

Choose Lagrangian multipliers 1, 1, 1,

Integrating, we get

Example 1.4b(14): Show that the integral surface of the equation 2y (z − 3) p + (2x − z) q = y (2x - 3) that passes through the circle x2 + y2 = 2x, z = 0 is x2 + y2 – z2 − 2x + 4z = 0.

Solution :

Given: 2y (z - 3) p + (2x − z) q = y (2x − 3)

This equation is of the form Pp + Qq = R

The required surface has to pass through

substitute for a and b from (2) and (3) in (9), we get

x2 + y2 - z2 - 2x + 4z = 0 which is the equation of the required integral surface.

EXERCISES 1.4.b

Find the general solution of