Answer
$y=\frac{2e^2}{e-1}(x-1)-2x^2 \ln (x)-\frac{2e^2}{e-1}$
Work Step by Step
Integrate the function to turn $y''$ into $y'$.
$$y''=-2(3+2\ln (x))$$
$$y'=-2(x+2x\ln (x))+C_1$$
Integrate once again to turn $y'$ into $y$.
$$y=x(C_1-2x\ln (x))+C_2$$
Solve for the initial values.
$$y(1)=(C_1-2\ln (1))+C_2=0$$
$$C_1+C_2=0$$
$$y(e)=e(C_1-2e\ln (e))+C_2$$
$$C_1e-2e^2+C_2=0$$
Then we have the system:
$C_1+C_2=0$
$C_1e-2e^2+C_2=0$
Using substitution:
$C_1=-C_2$
$C_1e-2e^2-C_1=0$
then
$C_1=\frac{2e^2}{e-1}$
$C_2=-\frac{2e^2}{e-1}$
Substitute these constant values to get $y=\frac{2e^2}{e-1}(x-1)-2x^2 \ln (x)-\frac{2e^2}{e-1}$
