# Thread: Problem finding integrating factor of simple first order ODE

1. ## Problem finding integrating factor of simple first order ODE

Hi ! So I'm having trouble with a problem asking to find the integrating factor of this equation :

$t^{2}(\frac{dx}{dt})+t^{5}x = 4t$

So my approach was to set the right-hand side to 0 to get a homogeneous DE and then use separation of variables.

This got me :

$\frac{dx}{dt} = -t^{5}x/t^{2} = -t^{3}x$

$\frac{dx}{x} = -t^{3}dt$

$x = ce^{\frac{-t^{4}}{4}}$

So the general solution to the inhomogeneous one with u as integrating factor is : x= $ue^{\frac{-t^{4}}{4}}$
But plugging this into the original equation I arrive at :

$u'=\frac{4e^{\frac{-t^{4}}{4}}}{t}$

Which I don't know how to integrate and doesn't look like it can be any of the answers I was offered (multiple-choice test).

Where did I go wrong ?

Thank you for your help !

2. ## Re: Problem finding integrating factor of simple first order ODE

You lost the 4t term when you moved everything to one side.

\displaystyle \begin{align*} t^2\,\frac{\mathrm{d}x}{\mathrm{d}t} + t^5\,x &= 4\,t \\ \frac{\mathrm{d}x}{\mathrm{d}t} + t^3\,x &= \frac{4}{t} \end{align*}

The integrating factor is \displaystyle \begin{align*} \mathrm{e}^{\int{t^3\,\mathrm{d}t}} = \mathrm{e}^{\frac{t^4}{4}} \end{align*} so we get

\displaystyle \begin{align*} \mathrm{e}^{\frac{t^4}{4}}\,\frac{\mathrm{d}x}{ \mathrm{d} t} + t^3\,\mathrm{e}^{\frac{t^4}{4}} \,x &= \frac{4\,\mathrm{e}^{\frac{t^4}{4}}}{t} \\ \frac{\mathrm{d}}{\mathrm{d}t} \left( \mathrm{e}^{\frac{t^4}{4}} \, x \right) &= \frac{4\,\mathrm{e}^{\frac{t^4}{4}}}{t} \\ \mathrm{e}^{\frac{t^4}{4}}\,x &= \int{\frac{4\,\mathrm{e}^{\frac{t^4}{4}}}{t}\, \mathrm{d} t } \\ x &= \mathrm{e}^{-\frac{t^4}{4}} \int{ \frac{4\,\mathrm{e}^{\frac{t^4}{4}}}{t}\,\mathrm{d }t } \end{align*}

This is the exact solution, although it can't be written explicitly in terms of elementary functions.

3. ## Re: Problem finding integrating factor of simple first order ODE

Thank you very much for your help !
I can't see where I missed the 4/t term ? Except setting it to 0 in the original equation at first to make it homogeneous.
Also is it not e^( -t^4/4) instead of t^4/4 ? Or did I make a mistake.

4. ## Re: Problem finding integrating factor of simple first order ODE

Another way of looking at it:
First, divide by $t^2$ to get $\frac{dx}{dt}+ t^3x= \frac{4}{t}$.

Now, by the definition of "integrating factor" we are looking for a function, u(t), such that $\frac{d(ux)}{dt}= u\frac{dx}{dt}+ \left(\frac{du}{dt}\right)x= u\frac{dx}{dt}+ t^3ux$. That is, u must satisfy the equation $\frac{du}{dt}= t^3u$. We can separate that into $\frac{du}{u}= t^3dt$. Integrating, $ln(u)= \frac{1}{4}t^4+ C$. Solving for u, $u= C'e^{(1/4)t^4}$ where $C'= e^C$. We only need one function so we can take C'= 1. Our integrating factor is $e^{(1/4)t^4}$.

Multiplying by that, the differential equation becomes $e^{(1/4)t^4}\frac{dx}{dt}+ t^3e^{(1/4)t^4}x= \frac{d}{dt}\left(e^{(1/4)t^4}x(t)\right)= \frac{e^{(1/4)t^4}}{t}$.

5. ## Re: Problem finding integrating factor of simple first order ODE

Originally Posted by Jo37
Thank you very much for your help !
I can't see where I missed the 4/t term ? Except setting it to 0 in the original equation at first to make it homogeneous.
Also is it not e^( -t^4/4) instead of t^4/4 ? Or did I make a mistake.
You made several mistakes. Instead of bleating about your own incorrect solution you should look through the correct solution given and try to determine where your mistakes are.