Converting Non-Linear DE: From Standard Form to Integrating Factor

[basty]

Homework Statement

This problem appears as one of exercise, of a linear differential equations chapter in my DE book.

$y \ dx - 4(x + y^6) \ dy = 0$

How to change above equation in a standard form of a linear differential equation?

The standard form of a linear differential equation is:

$\frac{dy}{dx} + P(x) y = f(x)$

It appears that the above equation is not linear because the power of $y$ is not 1 instead of 6.

Also, what is its integrating factor $(e^{\int P(x) \ dx})$?

Homework Equations

The Attempt at a Solution

$y \ dx - 4(x + y^6) \ dy = 0$
$y - 4(x + y^6) \frac{dy}{dx} = 0$
$4(x + y^6) \frac{dy}{dx} - y = 0$
$\frac{dy}{dx} - \frac{1}{4(x + y^6) }y = 0$

 

[LCKurtz]

Homework Statement

This problem appears as one of exercise, of a linear differential equations chapter in my DE book.

$y \ dx - 4(x + y^6) \ dy = 0$

How to change above equation in a standard form of a linear differential equation?

The standard form of a linear differential equation is:

$\frac{dy}{dx} + P(x) y = f(x)$

It appears that the above equation is not linear because the power of $y$ is not 1 instead of 6.

You are correct. It is not linear.

Also, what is its integrating factor $(e^{\int P(x) \ dx})$?

That's the wrong question to ask. Your equation is in the form $Pdx + Qdy=f(x)$. Look in your text about exact DE's and what condition on $P$ and $Q$ makes the equation exact.

[Edit, added] And if the equation isn't exact is it possible to find an integrating factor that makes it exact?