Question

Find the particular integral of

y' - y = 2e^x

Answer 1

Solution

General solution of homogeneous equation y' - y = 0 is y₀(x) = Ce^x.

So particular integral of y' - y = 2e^x

Is to be find as y(x) = Axe^x. Substitution into equation gives Ae^x + Axe^x - Axe^x = 2e^x  => A = 2

Therefore   

Answer

The particular integral of given equation is y(x) = 2xe^x