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Interpret the three conditions at zero as an initial-value problem on $x\ge 0$, with interior source $\xi >0$. A causal Green function satisfies ${\mathcal{L}}_{x}G=\delta (x-\xi )$ 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 $G^{\prime }$ continuous and $G^{\prime \prime }$ to jump by one. For $t=x-\xi >0$, the homogeneous roots are $0,2,-1$. Solving $g(0)=g^{\prime }(0)=0$, $g^{\prime \prime }(0)=1$ yieldsFor $\xi >0$ it vanishes near zero, so all three initial boundary conditions hold. The matching conditions ensure that ${\partial }_x^{3}G$ supplies precisely a Dirac delta, without unwanted delta derivatives.

The homogeneous solution carrying the nonzero initial data is $y_h=\frac{1}{2}-\frac{1}{6}{e}^{2x}+\frac{2}{3}{e}^{-x}$. The Green-function representation therefore givesFor $x<1$ the integral vanishes. For $x\ge 1$, set $t=x-1$ and integrate to obtain $\frac{e}{6}(e^{2t}-3e^t-e^{-t}+3)$. ThusThe bracket and its first two derivatives vanish at $x=1$, so $y,y^{\prime },y^{\prime \prime }$ 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.