# Particular solution / undetermined coefficients

1. Feb 1, 2008

### ssb

1. The problem statement, all variables and given/known data

tyʺ+yʹ=4t

2. Relevant equations

3. The attempt at a solution

The problem that I am having with this problem is ive never been shown how to calculate the particular solution when there is an unknown (t) on the left hand side of the equation. If the problem were y"+y'=4t then I would have no problem but that t in there is throwing me off and I don't know how to approach this problem. Can anyone offer me a hint?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 1, 2008

### ssb

heres an attempt:

tyʺ+yʹ=4t

Yp(t) = bt + a
Y'p(t) = b
Y"p(t) = 0

plugin to tyʺ+yʹ=4t

t(0) (hmm i may have answered my own question) + b + 0 (bt + a) = 4t

therefore b = 0 and a =0 (doesnt work)

Attempt 2

Yp(t) = at
Y'p(t) = a
Y"p(t) = 0

plug into tyʺ+yʹ=4t

t(0) + a + 0(at) = 4t
therefore a = 0 and this doesnt work!!! argg i am stuck

3. Feb 1, 2008

### Ben Niehoff

You're on the right track. Try re-writing it

$$y'' + \frac{1}{t}y' = 4$$

Now, try a more generic substitution:

$$y = a_n t^n$$

and see what happens. You should be able to determine what value(s) of n will make it possible to satisfy the equation.

4. Feb 1, 2008

### ssb

Thankyou for your hint. Ive attempted to use your hint and got stuck again. Here is what I did:

Yp(t) = At
Y'p(t) = A
Y"p(t) = 0

plug into $$y'' + \frac{1}{t}y' = 4$$

0 + $$\frac{1}{t}A = 4$$
therefore A=4t

plug that into yp(t) = (4t)t
but this is wrong as well!! Am I correct in what im doing when I solve for A and make it = to 4t?

Also I tried to begin with Yp(t) = At^2 and it didnt work as well.

5. Feb 1, 2008

### Ben Niehoff

Did you read the second part of my hint?

6. Feb 1, 2008

### ssb

so if I make
Yp(t) = aT
Y'p(t) = a
Y"p(t) = 0

then

0+(at)/t = 4
a=4

I understand the second hint you gave me but its just not clicking man. I tried a = 4 and it still didnt work. Do you have a suggestion for paticular solution I might try and calculate with? thanks

7. Feb 1, 2008

### Ben Niehoff

I don't know how much clearer to make it. You keep repeating your trial with n=1, even though you've already proven that doesn't work.

8. Feb 1, 2008

You sure?

9. Feb 1, 2008

### ssb

Thankyou so much!!! I finally figured it out. Thanks for nudging me in the proper direction. I really appreciate it (and i learned something new).

10. Feb 2, 2008

### HallsofIvy

Staff Emeritus
By the way, since this equation had y" and y" but not y itself, you it's easily reduced to a first order equation: let u= y' and then ty"+ ty'= tu'+ u= 4t. That's the same as u'+ u/t= 4 which is a linear first order equation and the integrating factor can be easily found.

Ben Niehoff's idea worked because if you multiply the equation by t, you get t2y"+ ty'= t2 in which each derivative is multiplied by a power of t equal to the degree of the derivative. That's called an "Euler type" or "Equi-potential" equation. "Trying" a solution of the form xn is like "trying" erx in an equation with constant coefficients. In fact, the change of variable x= ln u changes the equipotential equation into an equation with constant coefficients

Last edited: Feb 2, 2008