Nonhomogeneous Power Series Solution

[Sculptured]

For the fun of it, my DE book threw in a couple of problems involving nonhomogenous second order DE's in the section I'm currently going through. Although I have solved for the complementary solution, any suggestions on how to find the particular solution?

For example, the one I'm looking at right now is y''-xy = 1.

 

[HallsofIvy]

1= 1+ 0x+ 0x^2+ 0x^3+ \cdot\cdot\cdot
After you have written the left side of the equation as a power series in x, do the same on the right. Coefficients of powers on the left must equal coefficients of corresponding powers on the right. Here, your coefficient of x⁰ must be equal to 1 and all others equal to 0.

That was too easy. Suppose your equation were y"- xy= e^x?

Expand e^x in a power series:
e^x= 1+ x+ (1/2)x^3+ \cdot\cdot\cdot+ (1/n!)x^n+ \cdot\cdot\cdot[/itex]<br /> Now the coefficient of xⁿ, on the right, will be equal to 1/n!.

 

[Sculptured]

Thanks; makes good sense.

I have another question. This time it involves indicial roots. Working with the Frobenius method I find that in the DE: xy'' + 2y' -xy = 0 both of the indicial roots come to give me the same series solution. I used r = 0 to attain it and the book used r = -1. (or so I suppose since r=-1 is in their solution) The problem comes in that when I use r = 0 it doesn't solve the DE while r = -1 does. Is there any reason why there should be this difference between roots? Any rule of thumb to work with when picking which roots to use?

The book uses the larger root in its first solution when giving the general way of solving this case of the Frobenius method. When doing an actual example it uses the smaller root for the first solution. I take it that the smaller root should always be applied first?

 

[HallsofIvy]

To quote from a differential equations textbook,

If two roots of the indicial equation differ by an integer, it may be impossible to find an independent solution of the form
x^{\lambda_1}\sum_{n=0}^\infty{a_nx^n}
for the smaller root.

In that case, you can use the known solution, corresponding to the larger root of the indicial equation, to reduce the order of the equation and solve that for an independent solution. Of course, since you only know the first solution as an infinite series, that may be very difficult!

 

[matematikawan]

Thanks; makes good sense.

I have another question. This time it involves indicial roots. Working with the Frobenius method I find that in the DE: xy'' + 2y' -xy = 0 both of the indicial roots come to give me the same series solution. I used r = 0 to attain it and the book used r = -1. (or so I suppose since r=-1 is in their solution) The problem comes in that when I use r = 0 it doesn't solve the DE while r = -1 does. Is there any reason why there should be this difference between roots? Any rule of thumb to work with when picking which roots to use?

The book uses the larger root in its first solution when giving the general way of solving this case of the Frobenius method. When doing an actual example it uses the smaller root for the first solution. I take it that the smaller root should always be applied first?

I think you have got the indicial roots wrong. The correct roots are r=0 and r=1.

 

[cybla]

quick question. So for this question, y" - xy = 1...is the particular solution y=1?Also, how would i find the particular solution to y"+y=x?