Question

SOLVE THE GIVEN PROBLEMS BY LAGRANGE'S METHOD: a)2xzp +2yzp = z^2 - x^2 - y^2 b)(3-2yz)p + x(2z-1)q = 2x(y-3) c)x(z-3y^3)p + y(3x^3-z)q = 3(y^3-x^3)z d) (z^2 - 2yz - y^2)p + x(y+z)q = x(y-z) e)y^2 (x+y)p +x^2(x+y)q = (x^2 + y^2)z f)(x+2y^2+z)p +y(2x-1)q+2(x^2 +y^2+xz)=0 g) yp+xq = xyz^2(x^2-y^2) h)((x-y)y^2)p + ((y-x)x^2)q = (x^2 +y^2)z

Answer 1

Solving the questions using Lagrange's method we get the equations

Given:

The questions are given in differential form

To Find:

To get the equations i the form of partial differential equations

Solution:

The given equations are in differential form ,

To make them in the form of partial differential equations,

d) $(z^2 - 2yz - y^2)p + x(y+z)q = x(y-z)$

= $\frac{dx}{z^2 -2yz-y^2} = \frac{dy}{xy+zx} = \frac{dz}{xy-xz}$

Taking the last two functions ,

we get,

$\frac{dy}{xy+zx} = \frac{dz}{xy-xz}$

(y-z)dy = (y+z)dz

ydy - zdy = ydz +zdz

= ydy - zdy -ydz -zdz

ydy - d(zy) - zdz = 0

Upon Integrating on both the sides,

We get,

= $\frac{y^2}{2} -zy - \frac{z^2}{2}$ =c1

$y^2 - 2zy -z^2 =2c1=c2=u$

Using the Lagrange's Multipliers we get,

Each fractions,

xdx+ydy+zdz = 0

Upon integrating,

We get,

$\frac{x^2}{2} + \frac{y^2}{2} + \frac{z^2}{2}$ = c3

$x^2 +y^2 +z^2 =2c3=c4$

The required Solution is given as,

f($y^2 - 2zy -z^2$,$x^2 +y^2 +z^2$) = 0

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