Question

p cotx+ q cot y= cotz. solve by lagranges method DIFFERENTIAL EQUATIONS
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Answer 1

Answer:

for the Lagrange linear patial different equations p octx+ q cot y=cotz the general solution is your question

Answer 2

Given:

Lag ranges method in difference equation p cot x + q cot y =cot z.

To find:

The find p cot x + q cot y =cot z value

Step-by-step explanation:

p cot x + q cot y = cot z

the auxiliary equation is,

$\frac{dx}{cot x} =\frac{dy}{cot y} =\frac{dz}{cot z}$

Take first two,

∫$\frac{dx}{cot x} =$∫$\frac{dy}{cot y}$

∫tan x dx =∫tan y dy

log sec x =log sec y +log a

Taking last two,

∫$\frac{dy}{cot y}$ =∫$\frac{dz}{cot z}$

$\frac{cos z}{cos y} =b$

The general solution is,

Ф$(\frac{cos y}{cos x} ,\frac{cos z}{cos y})$ =0

Answer:

p cot x + q cot y = cot z is difference equation is Ф$(\frac{cos y}{cos x} ,\frac{cos z}{cos y})$=0.