The Student Room Group

Linear odes by contour integration

Can anyone show me the method for solving this:

xy'''-(1+x)y'-y=0

I'm using the substitution y(x)=(contour integral)[exp(xt)f(t)dt]
I manage to follow through with the working to get that:

f(t)=(t^-2)((t-1)^-2)((t+1)^-1)

Now I need to show that y=(contour integral)[(e^xt)f(t)dt] is a solution provided

[t(e^xt)/(t-1)]c. Where c is a contour to be evaluated.
How do I find closed form expressions for the solutions y1(x) and y2(x)?
Reply 1
MarioMaths
Can anyone show me the method for solving this:

xy'''-(1+x)y'-y=0

I'm using the substitution y(x)=(contour integral)[exp(xt)f(t)dt]
I manage to follow through with the working to get that:

f(t)=(t^-2)((t-1)^-2)((t+1)^-1) ) (*)

Now I need to show that y=(contour integral)[(e^xt)f(t)dt] is a solution provided

[t(e^xt)/(t-1)]c. Where c is a contour to be evaluated.
How do I find closed form expressions for the solutions y1(x) and y2(x)?


do you mean c is a constant to be evaluated?

surely the contour is given if you've been able to work out? in which case can you tell us what the contour is?
Reply 2
MarioMaths
Can anyone show me the method for solving this:

xy'''-(1+x)y'-y=0

I'm using the substitution y(x)=(contour integral)[exp(xt)f(t)dt]
I manage to follow through with the working to get that:

f(t)=(t^-2)((t-1)^-2)((t+1)^-1)

Now I need to show that y=(contour integral)[(e^xt)f(t)dt] is a solution provided

[t(e^xt)/(t-1)]c.
Where c is a contour to be evaluated.
How do I find closed form expressions for the solutions y1(x) and y2(x)?


What does this bold bit mean? Provided...? That seems to be an expression not an equation? :confused:
Reply 3
I think this might make things clearer.
Reply 4
I suspect the question wants you to find a specific contour for the boundary terms to drop off. For example, in this case shoving your ansatz in gives the requirement that the following integrand is zero:

[xt³f(t) - t(1+x)f(t) - f(t)]exp(xt)

Then do things like xf(t)exp(xt) = d/dt[ f(t)exp(xt) ] - f'(t)exp(xt) etc. so you can end up solving a fairly nice ODE in f(t) and are left to choose a contour such that the boundary terms give zero contributions. So you need a contour to start and finish at places where the boundary terms vanish, or choose a closed contour surrounding a some domain for which f(t)exp(xt) isn't analytic (otherwise this will generate a trivial solution). Check for independence by ensuring your choice of contours aren't equivalent upto some deformation.
Reply 5
But what exactly do I do?
I'm not completely understanding the process of finding such closed contours. How do I find them such that the solution for the differential equation is satisfied?
Reply 6
Well, assuming your calculations are correct (and that I understand your question), you need to find appropriate contours C1 and C2 such that:

t(t-1)-1ext (*)

is zero on dCi, i=1,2. So just choose your contours such that the endpoints lie at the zeros of (*), or a closed contour* that doesn't generate a trivial solution (i.e one that doesn't give yi=0). So at what points does (*) vanish in the (extended) complex plane? And is there a closed contour we could use that contains a simple pole (and hence generates a non-trivial solution)?

*This works because if Ci is closed dCi = Ø.
Reply 7
Well one solution should be the one I've stated above shouldn't it? It lies at the zeros of *.
Also, for the closed contour, could that just be any circle of radius centred at t=1?
Even if thats correct how would I present this solution?
Reply 8
Sorry, but I don't see where you've specified any contours?
Reply 9
the contour from 0 to -infinity
But I am not sure on the finite contours. I still am not sure what exactly what to do
Reply 10
How do I get (*) to vanish with other values?
What do I do to find the other contours c1 and c2?
Reply 11
Can anyone help in solving this?