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6.35 Finding the PI in exceptional cases

Suppose in the previous slide the equation was y′′+3⁢y′-4⁢y=2⁢exy^{\prime\prime}+3y^{\prime}-4y=2e^{x} instead of 2⁢e2⁢x2e^{2x}. We can’t have a PI of the form C⁢exCe^{x} since this is already a solution of the homogenous equation. In these circumstances we look for a PI of the form C⁢x⁢exCxe^{x}.

Method.

General principle If q⁢(x)q(x) is already a solution to the homogeneous equation then the PI should have the same form as q⁢(x)q(x), but multiplied by xx.

This explains the need for the special cases (“κ\kappa a single root” etc.) on slide 6.31. When κ\kappa is a double root of the auxiliary equation, not only eκ⁢xe^{\kappa x} but also x⁢eκ⁢xxe^{\kappa x} is a solution to the homogeneous equation and we look for a PI of the form C⁢x2⁢eκ⁢xCx^{2}e^{\kappa x}.

The same goes for polynomials: if 11 is a solution to the homogeneous equation then to find the PI for the case where q⁢(x)q(x) is a polynomial of degree nn, we consider polynomials of degree n+1n+1.