Answer
$y_p(x)=A_0e^{x}$
Work Step by Step
We are given $x^2y''+5xy'+7y=3e^{x}$
We have: $F(x)=3e^{x}$
Obtain: $3e^{x}(D-1)=0$
So, $D-1$ is the annihilator of $3e^{x}$
Therefore,
$(D-1)(x^2D^2+5xD+7)=0$
The trial solution is
$y_p(x)=A_0e^{x}$
