= Solution
Interpret the three conditions at zero as an initial-value problem on $x\ge0$, with interior source $\xi>0$. A <causal Green function> satisfies $\mathcal L_xG=\delta(x-\xi)$ and vanishes for $x<\xi$. Since the leading coefficient of the third derivative is one, the <jump conditions for a third-order Green function> require $G$ and $G'$ continuous and $G''$ to jump by one. For $t=x-\xi>0$, the homogeneous roots are $0,2,-1$. Solving $g(0)=g'(0)=0$, $g''(0)=1$ yields
$$
\boxed{G(x;\xi)=\Theta(x-\xi)\left(\frac{e^{2(x-\xi)}}6+\frac{e^{-(x-\xi)}}3-\frac12\right).}
$$
For $\xi>0$ it vanishes near zero, so all three initial boundary conditions hold. The matching conditions ensure that $\partial_x^3G$ supplies precisely a Dirac delta, without unwanted delta derivatives.
The homogeneous solution carrying the nonzero initial data is $y_h=\tfrac12-\tfrac16e^{2x}+\tfrac23e^{-x}$. The <Green-function representation> therefore gives
$$
y(x)=y_h(x)+\int_0^xG(x;\xi)e^\xi\Theta(\xi-1)\,d\xi.
$$
For $x<1$ the integral vanishes. For $x\ge1$, set $t=x-1$ and integrate to obtain $\tfrac e6(e^{2t}-3e^t-e^{-t}+3)$. Thus
$$
\boxed{y(x)=\frac12-\frac16e^{2x}+\frac23e^{-x}+\frac e6\Theta(x-1)\left(e^{2(x-1)}-3e^{x-1}-e^{-(x-1)}+3\right).}
$$
The bracket and its first two derivatives vanish at $x=1$, so $y,y',y''$ match continuously there; its third derivative jumps by the stated forcing. The value assigned to $\Theta(0)$ is immaterial. The source at the initial endpoint $\xi=0$ would need a separate endpoint-delta convention, which is not used in this integral.
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