Particular solution / undetermined coefficients

[ssb]

Homework Statement

tyʺ+yʹ=4t

Homework Equations

The Attempt at a Solution

The problem that I am having with this problem is I've 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?

 

[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 doesn't work! argg i am stuck

 

[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.

 

[ssb]

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.

Thankyou for your hint. I've 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 I am 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.

 

[Ben Niehoff]

Did you read the second part of my hint?

 

[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

 

[Ben Niehoff]

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.

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.

 

[Ben Niehoff]

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

You sure?

 

[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).

 

[HallsofIvy]

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" e^rx 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