Answer
$y(x)=C_1xe^{3x}+C_2e^{3x}+4x^{\frac{5}{2}}e^{3x}$
Work Step by Step
Given $y''-6y'+9y=15e^{3x}\sqrt x$
with $y_1=e^{3x}$
We first compute the appropriate derivatives
$y'=e^{3x}(u'+u)=e^{3x}u'+e^{3x}3u\\
y''=e^{3x}(u''+2u'-u)=e^{3x}u''+e^{3x}6u'+9e^{3x}u$
Substituting these expressions into the given equation, we find that $u$ must satisfy
$e^{3x}u''+e^{3x}6u'+9e^{3x}u-6(e^{3x}u'+e^{3x}3u)+9ue^{3x}=15e^{3x}\sqrt x$
which simplifies to
$e^{3x}u''=15e^{3x}\sqrt x$
or equivalently
$u''=15\sqrt x$
Then
$u'=10x^{\frac{3}{2}}+C_1$
where $C_1$ is a integration constant.
Obtain:
$u=4x^{\frac{5}{2}}+C_1x+C_2$
The general solution is
$y(x)=C_1xe^{3x}+C_2e^{3x}+4x^{\frac{5}{2}}e^{3x}$
