solve the following differential equation by the showing the exactness of the differential equation x(x^2+y^2-a^2 )dx+y(x^2-y^2-b^(2 ) )dy=0
Answers
Answer:
Exact Equation
An "exact" equation is where a first-order differential equation like this:
M(x, y)dx + N(x, y)dy = 0
has some special function I(x, y) whose partial derivatives can be put in place of M and N like this:
∂I∂x dx + ∂I∂y dy = 0
and our job is to find that magical function I(x, y) if it exists.
We can know at the start if it is an exact equation or not!
Imagine we do these further partial derivatives:
∂M∂y = ∂2I∂y ∂x
∂N∂x = ∂2I∂y ∂x
they end up the same! And so this will be true:
∂M∂y = ∂N∂x
When it is true we have an an "exact equation" and we can proceed.
And to discover I(x, y) we do EITHER:
I(x, y) = ∫M(x, y) dx (with x as an independent variable), OR
I(x, y) = ∫N(x, y) dy (with y as an independent variable)
And then there is some extra work (we will show you) to arrive at the general solution
Step-by-step explanation:
Exact Equation
An "exact" equation is where a first-order differential equation like this:
M(x, y)dx + N(x, y)dy = 0
has some special function I(x, y) whose partial derivatives can be put in place of M and N like this:
∂I∂x dx + ∂I∂y dy = 0
and our job is to find that magical function I(x, y) if it exists.
We can know at the start if it is an exact equation or not!
Imagine we do these further partial derivatives:
∂M∂y = ∂2I∂y ∂x
∂N∂x = ∂2I∂y ∂x
they end up the same! And so this will be true:
∂M∂y = ∂N∂x
When it is true we have an an "exact equation" and we can proceed.
And to discover I(x, y) we do EITHER:
I(x, y) = ∫M(x, y) dx (with x as an independent variable), OR
I(x, y) = ∫N(x, y) dy (with y as an independent variable)
And then there is some extra work (we will show you) to arrive at the general solution