Question

Find a particular solution to the differential equation:

y'' + 0y' - 1y = -1t^3

Answer 1

ok, for this one, I would use the method of undetermined coefficients

Answer 2

so it would take the form of A + Bt + Ct^2 + Dt^3

Answer 3

y=A+Bt+Ct^2+Dt^3
y'=B+2Ct+3Dt^2
y''=2C+6Dt
so we have 2C+6Dt-(A+Bt+Ct^2+Dt^3)=-1t^3

Answer 4

2C-A=0
6D-B=0
-C=0
-D=-1

Answer 5

If C=0, then 2(0)-A=0 implies A=0.
If D=1, then 6(1)-B=0 implies -B=-6 implies B=6.
So y=6t+t^3

Answer 6

gotta make dinner peace