Hello! Nonlinear Differential Equation

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Sep 2nd 2012, 02:25 PM
Nekoteko

Hello! 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 PM
HallsofIvy

Re: Hello! Nonlinear Differential Equation

?? That is a linear differential equation! It's integrating factor is easy to find.
Sep 2nd 2012, 03:35 PM
Nekoteko

Re: 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 PM
Prove It

Re: Hello! Nonlinear Differential Equation

If it can be rewritten as $\displaystyle \begin{align*} \frac{dy}{dx} + P(x)\,y = Q(x) \end{align*}$ , then you can easily find the integrating factor $\displaystyle \begin{align*} e^{\int{P(x)\,dx}} \end{align*}$ . How can you get your DE into this form?
Sep 3rd 2012, 07:27 AM
HallsofIvy

Re: Hello! Nonlinear Differential Equation

An integrating factor, for a linear differential equation, $f(x)dy/dx+ g(x)y= h(x)$ is a function, $\mu(x)$ , such that multiplying by it makes the left side an "exact derivative". That is, $f(x)\mu(x)dy/dx+ g(x)\mu(x)y= d(f(x)\mu(x)y)/dx$ . Theoretically, all first order differential equations have "integrating factors" but it is only linear equations for which it is easy to find because $d(f(x)\mu(x)y)/dx$ , by the product rule, is $f(x)\mu(x) dy/dx+ (f'(x)\mu(x)+ f(x)\mu'(x))y= f(x)\mu(x)dy/dx+ g(x)y$ so we must have $f'(x)\mu(x)+ f(x)\mu'(x)= \mu g(x)$ , a first order, separable, differential equation for $\mu$ . We can separate that as $f(x)\mu'(x)= \mu(x)(g(x)- f'(x)$ so that $\frac{d\mu}{\mu}= \frac{g(x)- f'(x)}{f(x)}dx$ , giving the exponential formula Prove It gave.

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

Re: 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.