Can you help find the implicit form of the equation:?

Question

Have to find the implict form of:

- (e^x+e^-x) / -(y (e^x-e^-x+1)^3 )

Sorry forgot the

dy/dx = -(e^x+e^-x) / -y(e^x-e^-x+1)^3

Answer 1

dy/dx = -(e^x + e^(-x)) / -(y (e^x - e^(-x) + 1)³)

So ∫ydy = ∫(e^x + e^(-x)) dx/(e^x - e^(-x) + 1)³

Let u = e^x - e^(-x) + 1 thus du = (e^x + e^(-x) ) dx

So ∫ydy = ∫du/u³

So ½y² = -½*1/u² + ½A

ie y² = A - 1/u²
= A - 1/(e^x - e^(-x) + 1)²

Answer 2

We are given
dy/dx = -(e^x+e^-x) / -y(e^x-e^-x+1)^3
So, its like
f(x) = dy/dx = -(e^x+e^-x) / -y(e^x-e^-x+1)^3

Derivate this w.r.t x
f'(x) = d(dy/dx)/dx =d[ -(e^x+e^-x) / -y(e^x-e^-x+1)^3]/dx
=>f'(x) = d[ -(e^x+e^-x) / -y(e^x-e^-x+1)^3]/dx
=>f'(x) = {[d[ -(e^x+e^-x)]/dx]*[-y(e^x-e^-x+1)^3 ]} / {[ -(e^x+e^-x)]*d[-y(e^x-e^-x+1)^3]//dx]}
[thia is because of the division rule of derivatives]
=>f'(x) = { [-(e^x - e^-x)]*[-y(e^x-e^-x+1)^3 ]} / {[-(e^x+e^-x)]*[-{[y'*(e^x-e^-x+1)^3] + y*3(e^x-e^-x+1)^2 * (e^x+e^-x)]}]}

RESONS FOR THIS STEPS
d[ -(e^x+e^-x)]/dx = -(e^x - e^-x)
d[-y(e^x-e^-x+1)^3]/dx = -{[y'*(e^x-e^-x+1)^3] + y*3(e^x-e^-x+1)^2 * (e^x+e^-x)]} [this is because of addition rule of derivatives]

=>f'(x) = { [y*(e^x - e^-x)]*[(e^x-e^-x+1)^3 ]} / { [e^x+e^-x]*{ [y'*(e^x-e^-x+1)^3] - 3y*(e^x+e^-x)*(e^x-e^-x+1)^2 }

=>d^2y/dx^2 - { [y*(e^x - e^-x)]*[(e^x-e^-x+1)^3 ]} / { [e^x+e^-x]*{ [dy/dx*(e^x-e^-x+1)^3] - 3y*(e^x+e^-x)*(e^x-e^-x+1)^2 } = 0

Solution of this equation will involve some constants and if you write this equation again in form of y=f(x), will give you implicit equation.

OR

dy/dx = -(e^x+e^-x) / -y(e^x-e^-x+1)^3
=>-y dy = [(e^x+e^-x) / (e^x-e^-x+1)^3] dx
Integrating on both the sides, will give

For integrating, right hand side
Let u = e^x-e^-x+1 => du/dx = e^x+e^-x
=> du = [e^x+e^-x]*dx

so we have
integration [1/u^3] du
=integration [u^(-3)] du
= -3u^(-4)
= -3[e^x-e^-x+1]^(-4)

and integrating left hand side we get, -(y^2)/2

So, we have
-(y^2)/2 = -3[e^x-e^-x+1]^(-4)
=>(y^2) = 6[e^x-e^-x+1]^(-4)
=>(y) = sqrt{6[e^x-e^-x+1]^(-4)}
=>y = sqrt(6) / [e^x-e^-x+1]^(2)

Even, this is the implicit form.

I hope I have answered your question.

All the best.

Answer 3

In the implicit form of a differential equation, the derivative dy/dx appears within the equation as y'; so your equation is already implicit:

y' = -(e^x +e^-x)/-y (e^x-e^-x+1)^3

but would more normally appear as

y'y(e^x-e^-x+1)^3=e^x +e^-x

Answer 4

this is already implicit, and it is not correct, the = sign is missing