# Simultaneous Linear Differential Equations

• ### Introduction

In the previous blospost, we covered the L.D.E.s of one dependent variable ${y}$ and one independent variable ${x}$. Moving ahead, we will now study L.D.E.s of two or more dependent variables and one independent variable.

In order to solve such equations, we will need as many differential equations as the number of dependent variables. This is similar to what we need when we solve simultaneous equations of the form ${2x+3y = 5, \ 5x - 4y = 6}$, 2 unknowns, hence 2 equations.

Note that a system of ${n}$ first-order linear differential equations with ${n}$ variables can be converted into a single differential linear equation of order ${n}$.

• ### The Method of Substitution/ Elimination (2 Independent Variables Only)

This method is similar to solving the linear equations simultaneously. One has to eliminate the other variable by some operation. Having done that, a L.D.E. with only 1 independent variable is obtained and can be solved by the methods learnt earlier.

Once one variable is obtained, the second one can be easily obtained.

• ### Symmetrical Simultaneous D.E.s

When the equations are of the form

${\frac {dx}{P} = \frac {dy}{Q} = \frac {dz}{R},}$

where ${P,Q}$ and ${R}$ are functions of ${x,y,z}$, they are said to be symmetrical simultaneous D.E.s. The solution consists of 2 independent relations,

${f_1 (x,y,z)= c_1 \ and \ f_2 (x,y,z) = c_2}$

• ### Method of Grouping

If in case it so happens that any 2 of the 3 equations above contain only 2 variables, they can be solved readily by variable-separable method.

• ### Method of Multipliers

We find a set of multipliers, say ${l,m}$ and ${n}$, not necessarily constants in such a way that,

${\frac {dx}{P} = \frac {dy}{Q} = \frac {dz}{R} = \frac {ldx + mdy + n dz}{lP + mQ + nR}}$

If, by choice, ${lP + mQ + nR =0}$, then

${l dx + m dy + ndz =0}$

This D.E. is now integrable. On integration, we get ${f_1 (x,y,z)= c_1}$.

We then choose another set of multipliers, follow the same procedure and get the 2nd relation.