Math, asked by krishnakumarid1969, 7 months ago

solve xdx/y²=dy/xz=dz/y²

Answers

Answered by aayushp2016
0

Answer:

oduction. Here we shall study the methods of solving simultaneous equations of the

first order and of the first degree in the derivatives. Here we consider equations involving

only three variables. The method of solution presented here can be applied to equations

involving any number of variables.

The general type of a set of simultaneous equations of the first order having three variables

is

P1dx + Q1dy + R1dz = 0, P2dx + Q2dy + R2dz = 0,

where the coefficients are functions of x, y, z. Solving these equations simultaneously, we

have

dx

Q1R2 − Q2R1

=

dy

R1P2 − R2P1

=

dz

P1Q2 − P2Q1

,

which is of the form

dx

P

=

dy

Q

=

dz

R

,

where P, Q, and R are functions of x, y, z. Thus we note that the simultaneous equations

(1) can always be put in the form (2).

Method - I for solving dx

P

=

dy

Q

=

dz

R

....(1)

By equating two of the three fractions of (1), we may be able to get an equation in

only two variables. Some times such an equation is obtained after cancelation of some

factor from the chosen two fractions of (1). On integrating the differential equation in

only two variables by well known methods, we shall obtain one of the relations in the

general solution of (1). This method may be repeated to give another relation with help

of two other fractions of (1).

Example. Solve dx

yz

=

dy

zx

=

dz

xy

.

Solution. Taking the first two fractions, we have

xdx = ydy or 2xdx − 2ydy = 0 so that x

2 − y

2 = c1.

Again, taking the first and the third fractions, we have

xdx = zdz or 2xdx − 2zdz = 0 so that x

2 − z

2 = c2.

Therefore the general solution is given by the relation x

2 − y

2 = c1 and x

2 − z

2 = c2.

Step-by-step explanation:

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