# Linear DE

## 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})##?

## 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 Helper
Gold Member

## 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?

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