#### Answer

False

#### Work Step by Step

$y_{p}=Ae^{x}$
$y_{p}'=Ae^{x}$
$y_{p}''=Ae^{x}$
Substitute equations in the differential equation to get
$(Ae^{x})-(Ae^{x})=e^{x}$
$0=e^{x}$
Which is not true, therefore the particular solution is not of the form $y_{p}=Ae^{x}$
Remember that if the sum of the coefficients of a differential equation is zero and $G(x)=e^{kx}P(x)$,
then $y_{p}$ is of the form
$xQ(x)e^{kx}$
Where P(x) and Q(x) are polynomials of the same degree
The sum of coefficients of the given differential equation is $1+0-1=0$
Therefore, the particular solution is of the form
$y_{p}=xAe^{x}$