Simple-looking 1st order DE, solution anyone?

[Irid]

I have a differential equation
<br /> \frac{dx}{dt} = \frac{a-x}{b-t} - cx<br />
but no idea if there is an analytical solution :(
I tried the integrating factor method, but I run into a e^(-x)/x^2 kind of integral, which I think doesn't have an analytical expression.. Help, please?

 

[ShayanJ]

With some manipulation,your equation can get the form below:
<br /> -(t-b)(\dot{x}+cx)+x=a<br />
Now take a solution like:
x=\sum_0^{\infty}a_n(t-b)^{n+r}
We have:
<br /> \dot{x}=\sum_0^{\infty}a_n (n+r) (t-b)^{n+r-1}<br />
If you substitute the x and \dot{x} in the homogenous DE,you will find:
<br /> -\sum_0^{\infty}a_n(n+r)(t-b)^{n+r}+\sum_0^{\infty}a_n(t-b)^{n+r}-\sum_1^{\infty}ca_{n-1}(t-b)^{n+r}=0<br />
Where I have changed the summation index in the third term from n to n-1 to make the powers of (t-b) the same in all summations.
Now to make the summations start from a common n,I get the n=0 terms out of the first two summations:
<br /> -a_0r(t-b)^r+a_0(t-b)^r=0<br />
Which gives r=1
Then you can take all the terms in one summation and equate the coefficient of (t-b)^{n+r} to zero to get a recursion relation which determines the series solution.
Then you should add a/(c+1) to get the solution to the in-homogenous equation.

 

[Irid]

so you mean to say that there is no solution in terms of elementary functions?

 

[ShayanJ]

The series you find may come out to be the power series corresponding to an elementary function.But it only may.

 

[Irid]

argh :(
better integrate numerically then...

 

[ShayanJ]

I just used www.wolframalpha.com to solve your equation and this is the result:
http://www3.wolframalpha.com/Calculate/MSP/MSP1811gi6c448044hadb10000242868gf692egc43?MSPStoreType=image/gif&s=12&w=389.&h=22.
With x \rightarrow t and y \rightarrow x
And Ei(x) is the exponential integral which you can find its values in some tables of functions.

 

[HallsofIvy]

That is a linear differential equation.
\frac{dx}{dt}+ (\frac{1}{b- t}+ c)x= \frac{a}{b- t}
An "integrating factor" is a function u(t) such that
\frac{dux}{dt}= u\frac{dx}{dt}+ u(\frac{1}{b- t}+ c)x
Since
\frac{dux}{dt}= u\frac{dx}{dt}+ \frac{du}{dt}x
that means we must have
\frac{du}{dt}= \frac{1}{b- t}+ c
du= \left(\frac{1}{b- t}+ c\right)dt

u(t)= -ln|b-t|+ ct

That is, the original differential equation is
\frac{d}{dt}\left(x(t)(-ln|b- t|+ c\right)= (-ln|b- t|+ ct)\frac{a}{b- t}
so that
x(t)(-ln|b- t|+ c)= \int (-ln|b-t|+ ct)\frac{a}{b- t} dt

Let u= b- t so that du= dt and the right side becomes
b\int \frac{ln|u|}{u}du- ac\int \frac{b- u}{u}du
In the first integral let v= ln|u| so that dv= du/u and that integral becomes
\int v dv= (1/2)v^2= (1/2)(ln|u|)^2= (1/2)(ln|b- t|)^2
In the second integral (b-u)/u= (b/u)- 1 and its integral bln|u|- u= b ln|b- t|- |b- t|.
putting those together
x(t)(-ln|b- t|+ c)= (1/2)(ln|b-t|)^2+ b ln|b- t|- |b- t|)

Finally,
x(t)= \frac{ln|b-t|)^2+ b ln|b- t|- |b- t|}{2(-ln|b- t|+ c)}

 

[Irid]

That is a linear differential equation.
\frac{dx}{dt}+ (\frac{1}{b- t}+ c)x= \frac{a}{b- t}
An "integrating factor" is a function u(t) such that
\frac{dux}{dt}= u\frac{dx}{dt}+ u(\frac{1}{b- t}+ c)x
Since
\frac{dux}{dt}= u\frac{dx}{dt}+ \frac{du}{dt}x
that means we must have
\frac{du}{dt}= \frac{1}{b- t}+ c
du= \left(\frac{1}{b- t}+ c\right)dt

u(t)= -ln|b-t|+ ct

That is, the original differential equation is
\frac{d}{dt}\left(x(t)(-ln|b- t|+ c\right)= (-ln|b- t|+ ct)\frac{a}{b- t}
so that
x(t)(-ln|b- t|+ c)= \int (-ln|b-t|+ ct)\frac{a}{b- t} dt

Let u= b- t so that du= dt and the right side becomes
b\int \frac{ln|u|}{u}du- ac\int \frac{b- u}{u}du
In the first integral let v= ln|u| so that dv= du/u and that integral becomes
\int v dv= (1/2)v^2= (1/2)(ln|u|)^2= (1/2)(ln|b- t|)^2
In the second integral (b-u)/u= (b/u)- 1 and its integral bln|u|- u= b ln|b- t|- |b- t|.
putting those together
x(t)(-ln|b- t|+ c)= (1/2)(ln|b-t|)^2+ b ln|b- t|- |b- t|)

Finally,
x(t)= \frac{ln|b-t|)^2+ b ln|b- t|- |b- t|}{2(-ln|b- t|+ c)}

This can't be right. If you set c=0, the solution is simply x= at/b, and your equation doesn't reduce to that. Also, if b -> infinity, the solution tends to x = Ae(-ct), which I see no trace of in your calculations.

 

[ShayanJ]

Here's the problem:

that means we must have
\frac{du}{dt}= \frac{1}{b- t}+ c
du= \left(\frac{1}{b- t}+ c\right)dt

u(t)= -ln|b-t|+ ct

We should have:
<br /> \frac{du}{dt}=u(\frac{1}{b-t}+c)<br />
Which gives an integrating factor of:
<br /> u=\frac{e^{ct}}{b-t}<br />
And we will have:
<br /> \frac{xe^{ct}}{b-t}=\int \frac{ae^{ct}}{(b-t)^2}dt<br />
And this gives the answer that I tried to present in the post #6 but it seems that its image is deleted from the server of the wolframalpha.
The answer is:
<br /> -{{\rm e}^{-ct}}a \left( -{{\rm e}^{ct}}+{{\rm e}^{cb}}{\it Ei}<br /> \left( 1,cb-ct \right) cb-{{\rm e}^{cb}}{\it Ei} \left( 1,cb-ct<br /> \right) ct \right) <br />
Where:
<br /> Ei(a,z)=\int_1^{\infty} e^{-tz}t^{-a}dt<br />
is the exponential integral and you can find its values in some mathematical tables.

 

[maistral]

Just trying :|

Isn't it a linear DE? I'm not really good at these kinds of things, but here's my try:

I'm not sure, I don't really go for analytic solutions :|EDIT: I looked at how wolframalpha would integrate the right-hand integral and it generated an Ei(-c(t-b)) o-o