Solving Nonhomogeneous Cauchy Equations with Erwin Kreyszig's WILEY Book

[Hussam Al-Tayeb]

I got this book from WILEY by Erwin Kreyszig. It tells how to solved homogenous cauchy equations. It also covers simple nonhomogenous equations.
But it doesn't cover when we have nonhomogenous Cauchy equations like this one.
x2y''-xy'+y=lnx
How do I go about solving that equation?

I substituted x=e ^t and obtained the homogeneous solution yh=c1*x+c2*x
but there is still the partial solution (yp).
Any idea?
final answer should be y=yh+yp

 

[rbzima]

I got this book from WILEY by Erwin Kreyszig. It tells how to solved homogenous cauchy equations. It also covers simple nonhomogenous equations.
But it doesn't cover when we have nonhomogenous Cauchy equations like this one.
x2y''-xy'+y=lnx
How do I go about solving that equation?

I substituted x=e ^t and obtained the homogeneous solution yh=c1*x+c2*x
but there is still the partial solution (yp).
Any idea?
final answer should be y=yh+yp

You could use a series solution to this one. Although it might take some time, it might also clarify why the solution is what you say it is.

 

[HallsofIvy]

I got this book from WILEY by Erwin Kreyszig. It tells how to solved homogenous cauchy equations. It also covers simple nonhomogenous equations.
But it doesn't cover when we have nonhomogenous Cauchy equations like this one.
x2y''-xy'+y=lnx
How do I go about solving that equation?

I substituted x=e ^t and obtained the homogeneous solution yh=c1*x+c2*x
but there is still the partial solution (yp).
Any idea?
final answer should be y=yh+yp

Well, no, you didn't get that as a homogeneous solutions because you obviously (unless you made a typo) don't have two independent solutions! If you make the substitution
x= e^t then you get the homogeneous equation d^2y/dt^2- 2dy/dt+ y= 0 which has characteristic equation r^2- 2r+ 1= (r-1)^2= 0 and so gives the single solution y= e^t. The OTHER independent solution is y= te^t. Those two solutions, converted to x, since t= ln(x) are y= x and y= x ln(x).

But why stop at the homogeneous equation. Since the right hand side of your equation is ln(x), replacing x by e^t there gives ln(e^t)= t. Your equation reduces completely to
d^2/dt^2- 2dy/dt+ y= t. Trying y= At+ b, y'= A, y"= 0 so you have -2A+ At+ B= t: A= 1 and -2A+ B= -2+ B= 1 gives B= 3. Your general solution to the converted equation is
y= C1e^t+ C2xe^t+ t+ 3 which, using t= ln x, goes back to y= C1x+ C2x ln(x)+ ln(x)+ 3.

 

[Hussam Al-Tayeb]

HallsofIvy, I understood the y1= e^t and y2=te^t

But you said "Your equation reduces completely to d^2/dt^2- 2dy/dt+ y= t"
shouldn't it reduce to:
e^2t y'' - e^t y' + y = t ?