Particular solution / undetermined coefficients

In summary, Ben Niehoff's idea for solving the equation y''+ y'=4t using a substitution worked because it transformed the equation into an equation with constant coefficients.
  • #1
ssb
119
0

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?
 
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  • #2
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
 
  • #3
You're on the right track. Try re-writing it

[tex]y'' + \frac{1}{t}y' = 4[/tex]

Now, try a more generic substitution:

[tex]y = a_n t^n[/tex]

and see what happens. You should be able to determine what value(s) of n will make it possible to satisfy the equation.
 
  • #4
Ben Niehoff said:
You're on the right track. Try re-writing it

[tex]y'' + \frac{1}{t}y' = 4[/tex]

Now, try a more generic substitution:

[tex]y = a_n t^n[/tex]

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 [tex]y'' + \frac{1}{t}y' = 4[/tex]

0 + [tex]\frac{1}{t}A = 4[/tex]
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.
 
  • #5
Did you read the second part of my hint?
 
  • #6
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
Ben Niehoff said:
Now, try a more generic substitution:

[tex]y = a_n t^n[/tex]

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.
 
  • #8
ssb said:
Also I tried to begin with Yp(t) = At^2 and it didnt work as well.

You sure?
 
  • #9
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
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
 
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Related to Particular solution / undetermined coefficients

1. What is a particular solution in the context of undetermined coefficients?

A particular solution, also known as a particular integral, refers to a specific solution to a nonhomogeneous linear differential equation that satisfies both the differential equation and any initial conditions given. It is found using the method of undetermined coefficients.

2. What is the method of undetermined coefficients?

The method of undetermined coefficients is a technique used to find a particular solution to a nonhomogeneous linear differential equation. It involves assuming a particular form for the solution and determining the coefficients based on the differential equation and any initial conditions given.

3. When is the method of undetermined coefficients applicable?

The method of undetermined coefficients is applicable to nonhomogeneous linear differential equations with constant coefficients and a right-hand side that consists of a linear combination of known functions such as polynomials, exponential functions, and trigonometric functions.

4. What is the difference between a homogeneous and nonhomogeneous linear differential equation?

A homogeneous linear differential equation has a right-hand side of zero, meaning there are no known functions present. A nonhomogeneous linear differential equation has a right-hand side that consists of known functions, making it more complex to solve.

5. How can I determine the undetermined coefficients for a particular solution?

To determine the undetermined coefficients, you must first assume a particular form for the solution and then substitute it into the original differential equation. This will result in a system of equations that can be solved to find the values of the coefficients. The values of the coefficients can also be determined by using the method of undetermined coefficients table for common functions.

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