Finding general solutions of forced equations

[metalscot]

I am having trouble after finding the complimentary solution to this problem, if possible I would like some guidance on how to proceed with the particular solution

Find a general solution to the following equation: x'' + 2x' + x = 3


x'' + 2x' + x = 3

complimentary solution when = 0

x'' + 2x' + x = 0

m^2 + 2m + 1 = 0

so this has only one root; m=-1

complimentary solution must then be in the form

x(t) = Ae^mt + Bte^mt

particular solution when = 3

x'' + 2x' + x = 3

From this point I am unsure how to proceed and would value any help

 

[LCKurtz]

Look for a solution of the form y_p = C, a constant.

 

[metalscot]

Look for a solution of the form y_p = C, a constant.

Thank you LCKurtz, in this example because I have x''+2x'+x=3

If I plug a value of x=3 into my original equation I end up with x=3 as a final answer so

x(t)particular=3

giving me x(t)general= x(t)complimentary + x(t)particular:

x(t)=Ae^-1t + Bte^-1t + 3 ?

 

[Hootenanny]

Thank you LCKurtz, in this example because I have x''+2x'+x=3

If I plug a value of x=3 into my original equation I end up with x=3 as a final answer so

x(t)particular=3

giving me x(t)general= x(t)complimentary + x(t)particular:

x(t)=Ae^-1t + Bte^-1t + 3 ?

Spot on

There are some general "rules" when finding particular solutions to ODEs, depending on the form of the forcing term. You will likely find them in your textbook.

 

[metalscot]

Spot on

There are some general "rules" when finding particular solutions to ODEs, depending on the form of the forcing term. You will likely find them in your textbook.

Thanks Hootenanny, I have a more complex forcing term to move onto now but with this basis I think I will get there.