You made two major errors. First, you found the antiderivative only of the exponential term and forgot all about the $-x$ term: you should have got $y=\frac12e^{2x}-\frac{x^2}2+c\;.$ Note that this is $y$, not $dy/dx$.
Secondly, you found all of the antiderivatives, but the problem requires you to find the specific antiderivative such that $y=5$ when $x=0$. Take the corrected general antiderivative,
$y=\frac12e^{2x}-\frac{x^2}2+c\;,$
and substitute $y=5$ and $x=0$ to get
$5=\frac12e^0-\frac02+c=\frac12+c\;;$
clearly we must have $c=5-\frac12=\frac92\;,$ and the unique antiderivative that satisfies the extra condition is
$y=\frac12e^{2x}-\frac{x^2}2+\frac92=\frac12\left(e^{2x}-x^2+9\right)\;.$