This is a good example of a question to which one can answer at some very different levels of mathematical sophistication... Since you say nothing about this, let me try an elementary approach.
What you call the Dirac delta function (which is not a function, at least not in the sense of a function from $\mathbb R$ to $\mathbb R$) is a strange object but something about it is clear:
One asks that $\displaystyle\int_y^z\delta(x)\mathrm dx=0$ if $y\leqslant z<0$ or if $0<y\leqslant z$ and that $\displaystyle\int_y^z\delta(x)\mathrm dx=1$ is $y<0<z$.
We will not use anything else about the Dirac $\delta$.
If one also asks that $\displaystyle\int_y^zu''(x)\mathrm dx=u'(z)-u'(y)$ for every $y\leqslant z$, one can integrate once your equation $\color{red}{-u''=\delta}$, getting
that there exists $a$ such that
$$
u'(x)=a-[x\geqslant0],
$$
where we used Iverson bracket notation. Now let us integrate this once again.
Using the facts that $\displaystyle\int_y^zu'(x)\mathrm dx$ should be $u(z)-u(y)$ for every $y\leqslant z$, and the value of $\displaystyle\int_y^z[x\geqslant0]\mathrm dx$, one gets that for every fixed negative number $x_0$,
$$
u(x)=u(x_0)+a\cdot (x-x_0)-x\cdot[x\geqslant0].
$$
This means that $b=u(x_0)-a\cdot x_0$ does not depend on $x_0<0$, hence finally, for every $x$ in $\mathbb R$,
$$
\color{red}{u(x)=a\cdot x+b-x\cdot[x\geqslant0]}.
$$
(And, in the present case, the condition that $u(-2)=u(3)=0$ imposes that $a=3/5$ and $b=6/5$.)
This is the general solution of the equation $-u''=\delta$. Note that every solution $u$ is $C^\infty$ on $\mathbb R\setminus\{0\}$ but only $C^0$ at $0$ hence $u'$ and $u''$ do not exist in the rigorous sense usually meant in mathematics. Note finally that $u$ is also
$$
u(x)=a\cdot x+b-x\cdot[x\gt0].
$$
A start: That it is a solution can be verified by differentiating the given function twice (use the Fundamental Theorem of Calculus). We also need to check that the initial conditions are satisfied. The calculations are straightforward.
For finding a solution without being told what it is, the standard trick is to multiply through by $y'$. Then integrate. We get $(y')^2=y^3+C$. Continue.
Best Answer
$Re(x)$ gives you the real part: $Re(x) = Re(a + bi) = cos(a)$
$Im(x)$ gives you the imagniary part: $Im(x) = Re(a + bi) = sin(b)$
In your example: $x_P = Re(z_p) = Re(\frac{te^{-t}( cos(t) + i sin(t) )}{2i}) = Re(\frac{te^{-t}cos(t)}{2i} + \frac{te^{-t}sin(t)}{2}) = Re(-\frac{ite^{-t}cos(t)}{2} + \frac{te^{-t}sin(t)}{2}) = \frac{te^{-t}sin(t)}{2}$
$Im(z_p) = Im(-\frac{ite^{-t}cos(t)}{2} + \frac{te^{-t}sin(t)}{2}) = -\frac{te^{-t}cos(t)}{2}$.