Linear, nonhomogenous, 2nd order ODE IVP

  • #1

Homework Statement


y''+4y=t^2+3e^t
y(0)=0
y'(0)=2

Homework Equations


CE: r^2+4
r=+/-2i
gs: y=c1 cos(2t) + c2 sin(2t)

The Attempt at a Solution



guess:
yp=(At^2+Bt+C)e^t
yp'=At^2e^t+2Ate^t+Bte^t+Be^t+Ce^t
yp''=At^2e^t+4Ate^t+Bte^t+2Ae^t+2Be^t+Ce^t

back into problem:
At^2e^t+4Ate^t+Bte^t+2Ae^t+2Be^t+Ce^t+4At^2e^t+4Ce^t=t^2+3e^t

which becomes:
5At^2e^t=t^2 then becomes: A=e^-t/5
then
4At^et+Bte^t+4Bte^t=0 which becomes: 4A+B+4B=0, B=-(4e^-t)/25

then:
2Ae^t+2Be^t+Ce^t+4Ce^t=3e^t, e^t's cancel which becomes:
2A+2B+5C=3 ---> C=[(75e^t)-2)e^-t]/125

ok, i'm sure i'm not doing this correctly. the book answer is:

7/10sin(2t)-19/40cos2t+1/4t^2-1/8+3/5e^t

what am i doing wrong? am i skipping a step?
 

Answers and Replies

  • #2
78
0
do you know the annihilator method? Don't waste your time with this guesswork and learn the annihilator method, it will save you from unnecessary stress.
 
  • #3
When you're working out the particular solution, deal with each term on the right hand side SEPERATELY.
First consider = t^2, then consider = 3e^t, otherwise you'll confuse yourself.
i.e. do what you've been doing, except consider them as two seperate cases, and then combine them in the final step.
Hope that helps.
 
Last edited:
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
962
Your "guess" is wrong. You dont' have t2e-t on the right hand side, you have t2+ e-t. You should be trying At2+ Bt+ C+ Ee-t.

Or, as Pseudo Statistic said, handle them separately: try At2+ Bt+ C for t2 and Ee-t for e-t.
 
  • #5
oh. ok let me try that. is separating them called superposition?

i don't know the annihilator method, but i'll do some research to find out...would appreciate it if you could explain it if it's right off the top of your head. thanks!
 
  • #6
ok this is what i did:
CE: r^2+4
r=+/-2i
gs: y=c1 cos(2t) + c2 sin(2t)

separating: y''+4y=t^2
yp=At^2+Bt+C
yp'=2At+B
yp''=2A becomes 1/4t^2

y''+4y=3e^t
yp=yp'=yp''=Ae^t
solve and get 3/5e^t

so total is:
C2sin(2t)+C1cos2t+1/4t^2+3/5e^t
pluggin in the IC i get:

7/10sin(2t)+3/5cos2t+1/4t^2+3/5e^t

did i do this correctly? how come i'm not gettin the book answer? i looked through my algebra and it does cancel out to be c1=3/5. am i forgetting to add a constant somewhere?
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,833
962
ok this is what i did:
CE: r^2+4
r=+/-2i
gs: y=c1 cos(2t) + c2 sin(2t)

separating: y''+4y=t^2
yp=At^2+Bt+C
yp'=2At+B
yp''=2A becomes 1/4t^2
WHAT becomes 1/4t^2? You want to solve y"+ 4y= t^2. So far you have yp"= 2A and 4yp= 4(At^2+ Bt+ C)= 4At^2+ 4Bt+ 4C so
yp"+ 4yp= 4At^2+ 4Bt+ (2A+4C)= t^2. For that to be true for all t, you must have 4A= 1, 4B= 0, 2A+ 4C= 0. C is not 0.

y''+4y=3e^t
yp=yp'=yp''=Ae^t
solve and get 3/5e^t
This is correct- you remembered the "4yp"!

so total is:
C2sin(2t)+C1cos2t+1/4t^2+3/5e^t
No. Your t^2 part is wrong.

pluggin in the IC i get:

7/10sin(2t)+3/5cos2t+1/4t^2+3/5e^t

did i do this correctly? how come i'm not gettin the book answer? i looked through my algebra and it does cancel out to be c1=3/5. am i forgetting to add a constant somewhere?
 
  • #8
oh sorry. i meant after the algebra, A becomes 1/4t^2. i guess maybe my entire mistake is my assumption that C=0. thanks! let me try it again.
 

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