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Second Order Non Homogeneous DE

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  • #1
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Homework Statement



Find the solution to :y''+2y'+y=t

Homework Equations



Suppose y(t)=B1t2+B2t+B3

And I believe, Y(t)=Yh+Yp. That is the solution is equal to the solution to the homogenous equation, plus the particular solution.

The Attempt at a Solution


First Solve the homogeneous equation:
y(t)=B1t2+B2t+B3
y'(t)=2B1t+B2
y''(t)=2B1

Substitute this back into the homogenous problem(y''+2y'+y=0) gives:

2B1+2(2B1t+B2)+B1t2+B2t+B3=0

Rearrange:

B1(t2+4t+2) +B2(t+2)+B3=0

Solution is t=-2 or B1,B2,B3=0

Now I'm unsure of what to do?

As for the particular solution say Yp:

Yp=ct, where c is a constant so,
Y'p=c
Y''p=0

Substitute this into :y''+2y'+y=t
0+2c+ct=t

c=?

Thanks for any help.
 

Answers and Replies

  • #2
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You should try using the annihilator method.

I got Yp=At+B
Yp'=A
Yp"=0 (A and B are constants)

plugging them back into the ODE :

2[A] + [At+B] = t

rearranging:

(2A+B)[1] + A[t] = t

so

A=1

2A+B=0


getting A=1 and B= -2

so then Yp= t-2

add it to Yh and you should get Y(t)
 
  • #3
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oh, and to find Yh just set the ODE = 0 and find the roots
 
  • #4
LCKurtz
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Homework Statement



Find the solution to :y''+2y'+y=t

Homework Equations



Suppose y(t)=B1t2+B2t+B3

And I believe, Y(t)=Yh+Yp. That is the solution is equal to the solution to the homogenous equation, plus the particular solution.

The Attempt at a Solution


First Solve the homogeneous equation:
y(t)=B1t2+B2t+B3
y'(t)=2B1t+B2
y''(t)=2B1
That is not the correct thing to try for the homogeneous solution. Try [itex]y = e^{rt}[/itex].
 
  • #5
HallsofIvy
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Homework Helper
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Homework Statement



Find the solution to :y''+2y'+y=t

Homework Equations



Suppose y(t)=B1t2+B2t+B3

And I believe, Y(t)=Yh+Yp. That is the solution is equal to the solution to the homogenous equation, plus the particular solution.

The Attempt at a Solution


First Solve the homogeneous equation:
y(t)=B1t2+B2t+B3
y'(t)=2B1t+B2
y''(t)=2B1
No, that is NOT how you solve for Yh.

Substitute this back into the homogenous problem(y''+2y'+y=0) gives:

2B1+2(2B1t+B2)+B1t2+B2t+B3=0

Rearrange:

B1(t2+4t+2) +B2(t+2)+B3=0

Solution is t=-2 or B1,B2,B3=0

Now I'm unsure of what to do?
That's because [itex]B1x^2+ B2x+ B3[/itex] is NOT a solution to the homogeneous equation. Use [itex]y= e^rt[/itex], as LCKurtz suggested, to get the "characteristic equation" to solve for r.

As for the particular solution say Yp:

Yp=ct, where c is a constant so,
Y'p=c
Y''p=0

Substitute this into :y''+2y'+y=t
0+2c+ct=t

c=?

Thanks for any help.
You cannot just let yp= ct. If the right hand side is a power of t, yp can be a polynomial nof degree at most that power- try yp= ct+ d.
 
  • #6
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Actually the annihilator method is correct. In fact I checked with my professor on this. The annihilator method takes out the factor of "guessing" a solution for the particular solution.

I even had my professor solve this and she got the same answer as I did, especially since we just went over this section in class.

Annihilator method is actually quite simple for this problem.
 
  • #7
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As for Yh, what I mean by set the ODE to 0 is:
y''+2y'+y=0

Let y''=r^2, y'=r, y=constant

ending up with:
r^2+2r+1=0

Factoring, you should end up with
(r+1)(r+1)=0

where r=-1,-1

Since this is a repeated root you should use:
Yh= c1*e^(rt) + c2*t*e^(rt)

Once you have Yh, then find Yp (particular solution) and add it to Yh for your general solution Y.

Just remember, since you DON'T have initial conditions

(for example: y'(0)=0, y(0)=1)

Your general solution is going to have constants c1 and c2 since there are no initial conditions given
 
  • #8
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Sorry for the late reply, yes I made a bad mistake with respect to the characteristic equation . I got it now though thanks!

Edit: I misunderstood the "hint" on the assignment. Every second order DE of this form, will have that solution given a single root.
 

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