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!

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- Sep 2nd 2012, 02:25 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, 03:24 PMHallsofIvyRe: Hello! Nonlinear Differential Equation
?? That is a

**linear**differential equation! It's integrating factor is easy to find. - Sep 2nd 2012, 03:35 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, 07:53 PMProve ItRe: Hello! Nonlinear Differential Equation
If it can be rewritten as $\displaystyle \displaystyle \begin{align*} \frac{dy}{dx} + P(x)\,y = Q(x) \end{align*}$, then you can easily find the integrating factor $\displaystyle \displaystyle \begin{align*} e^{\int{P(x)\,dx}} \end{align*}$. How can you get your DE into this form?

- Sep 3rd 2012, 07:27 AMHallsofIvyRe: Hello! Nonlinear Differential Equation
An integrating factor, for a linear differential equation, $\displaystyle f(x)dy/dx+ g(x)y= h(x)$ is a function, $\displaystyle \mu(x)$, such that multiplying by it makes the left side an "exact derivative". That is, $\displaystyle f(x)\mu(x)dy/dx+ g(x)\mu(x)y= d(f(x)\mu(x)y)/dx$. Theoretically,

**all**first order differential equations**have**"integrating factors" but it is only linear equations for which it is easy to find because $\displaystyle d(f(x)\mu(x)y)/dx$, by the product rule, is $\displaystyle f(x)\mu(x) dy/dx+ (f'(x)\mu(x)+ f(x)\mu'(x))y= f(x)\mu(x)dy/dx+ g(x)y$ so we must have $\displaystyle f'(x)\mu(x)+ f(x)\mu'(x)= \mu g(x)$, a first order,**separable**, differential equation for $\displaystyle \mu$. We can separate that as $\displaystyle f(x)\mu'(x)= \mu(x)(g(x)- f'(x)$ so that $\displaystyle \frac{d\mu}{\mu}= \frac{g(x)- f'(x)}{f(x)}dx$, giving the exponential formula Prove It gave.

In this case, $\displaystyle f(x)= x$ so $\displaystyle f'(x)= 1$ and $\displaystyle g(x)= -1$ so that equation is just $\displaystyle \mu+ x\mu'= -\mu$ which is the same as $\displaystyle x\mu'= -2\mu$ so that we have $\displaystyle \frac{d\mu}{\mu}= -\frac{2}{x}dx$. That's easy to integrate: $\displaystyle ln(\mu)= -2ln(x)+ C$. Since we only need one integrating factor, we can take C= 0 and then take the exponential of both sides to get $\displaystyle \mu= e^{-2ln(x)}= e^{ln(x^{-2})}= x^{-2}$ as integrating factor. - Sep 3rd 2012, 09:54 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.