This table is meant to clarify Table 3.6.1 in your textbook. Please let me know if you find any errors, or if you have a suggestion for making it clearer.

Forcing Function Form of y_p

t At+B

t² At²+Bt+C

t³ At³+Bt²+Ct+D

Etc. Etc.

sin(at) A sin(at) + B cos(at)

cos(at) A sin(at) + B cos(at)

But if sin(at) or cos(at) is a solution of the homogeneous problem instead use
y_p = A t sin(at) + B t cos(at)

t sin(at) (A t +B)sin(at) + (Ct+D) cos(at)

t cos(at) (A t +B)sin(at) + (Ct+D) cos(at)

But if t sin(at) or t cos(at) is a solution of the homogeneous problem instead use
y_p = (At²+Bt) sin(at) + (Ct²+Dt) cos(at)
This case cannot happen for 2nd problems, but can for higher order ones.

Etc. Etc.

e^at A e^at

But if e^at is a solution of the homogeneous problem instead use A t e^at. However, if
t e^at is also solution of the homogeneous problem then use
y_p = A t² e^at

t e^at (A t +B)e^at

But if e^at or t e^at is a solution of the homogeneous problem instead use
y_p = (A t² + Bt)e^at or, y_p = (A t³ + Bt²)e^at respectively.

t² e^at (At² +Bt + C)e^at

But if e^at or t e^at is a solution of the homogeneous problem instead use
y_p = (At³ +Bt² + Ct)e^at or, y_p = (At⁴ +Bt³ + Ct²)e^at respectively.

Etc. Etc.

e^atsin(bt) or e^atcos(bt) A e^atsin(bt) + B e^atcos(bt)

But if either is a solution of the homogeneous problem instead use
y_p = At e^atsin(bt) + Bt e^atcos(bt).

te^atsin(bt) or te^atcos(bt) (At+B) e^atsin(bt) + (Ct+D) e^atcos(bt)

But if e^atsin(bt) or e^atcos(bt) is a solution of the homogeneous problem instead use
y_p = (At²+Bt) e^atsin(bt) + (Ct²+Dt) e^atcos(bt)

Etc. Etc.