Hello, new here. Really need help with this one nonlinear differential equation problem from my Calc IV homework. :( Any help or hints would be greatly appreciated!

xy' - y = 1/(x^3)

Thank you in advance!

Printable View

- Sep 2nd 2012, 03:25 PMNekotekoHello! Nonlinear Differential Equation
Hello, new here. Really need help with this one nonlinear differential equation problem from my Calc IV homework. :( Any help or hints would be greatly appreciated!

xy' - y = 1/(x^3)

Thank you in advance! - Sep 2nd 2012, 04:24 PMHallsofIvyRe: Hello! Nonlinear Differential Equation
?? That is a

**linear**differential equation! It's integrating factor is easy to find. - Sep 2nd 2012, 04:35 PMNekotekoRe: Hello! Nonlinear Differential Equation
lolol Then I clearly know less than I thought.

I tried the integrating factor and got a little bit confused, but that's enough of a hint I guess. Thank you! Will try again & post results. - Sep 2nd 2012, 08:53 PMProve ItRe: Hello! Nonlinear Differential Equation
If it can be rewritten as , then you can easily find the integrating factor . How can you get your DE into this form?

- Sep 3rd 2012, 08:27 AMHallsofIvyRe: Hello! Nonlinear Differential Equation
An integrating factor, for a linear differential equation, is a function, , such that multiplying by it makes the left side an "exact derivative". That is, . Theoretically,

**all**first order differential equations**have**"integrating factors" but it is only linear equations for which it is easy to find because , by the product rule, is so we must have , a first order,**separable**, differential equation for . We can separate that as so that , giving the exponential formula Prove It gave.

In this case, so and so that equation is just which is the same as so that we have . That's easy to integrate: . Since we only need one integrating factor, we can take C= 0 and then take the exponential of both sides to get as integrating factor. - Sep 3rd 2012, 10:54 AMNekotekoRe: Hello! Nonlinear Differential Equation
Thanks, Prove It and HallsofIvy! Those equations really helped. I wish my textbook was as thorough as that post was.

I worked the integrating factor out and got x^-2, multiplied both sides of the equation and the general solution came out to be -3(x^-2) + Cx.