Solution of SDE with discontinuous coefficients

Question

The problem is stated as follows: $$dx = -k \, \text{sgn}(x) +dw(t),\\x(0)=y, \quad k>0,$$ where $w(t)$ is a standart wiener process. In the Caughey's paper the author asserts that applying Laplace transforming to Fokker-Planck equation with respect to $t$ and piecing together solutions of the resulting ODEs one can acquire the next solution: $$ p(x,t |y)=\frac{2л}{\sqrt\pi} e^{-2k|x|}\int_{(y+|x|)2\sqrt{t}}^{\infty} e^{-u^2}du+\frac{1}{\sqrt{2\pi t}}exp\left[k(x-|x|)-\frac{(x-y-kt)^2}{2t} \right]. $$ So the question is how has he executed that transformations? Fokker-Planck equation for this equation is written as $$ \frac{\partial p}{\partial t}=k \frac{\partial (sgn(x)p)}{\partial x}+\frac{\partial ^2 p}{\partial x^2}, $$ With initial condition $$ p(x,0)=\delta(x-y) $$ and boundary condition $$ \lim\limits_{|x|\rightarrow \infty}p(x,t)=0. $$ For the domain $x>0$ applying Laplace transformation we get $$ s \hat{p}(x,s)=k\frac{\partial \hat{p}}{\partial x} +\frac{\partial^2 \hat{p}}{\partial x^2} + \delta(x-y) $$ Assume we apply Laplace transformation again with respect to $x$ (let y>0). Denote the result $\varphi(m,s)=\mathcal{L}(p(x,s),m)$: $$ m^2 \varphi(m,s) + k m \varphi(m,s) -s \varphi(m) -e^{-y m} =0 $$ Inverting this expression we get $$ \hat{p}(x,s)=\frac{e^{(-\frac{k}{2}-\frac{\sqrt{k^2+4s}}{2})(x-y)} - e^{(-\frac{k}{2}+\frac{\sqrt{k^2+4s}}{2})(x-y)}}{\sqrt{k^2+4 s}} H(x-y), $$ where $H(x-y)$ is a Heaviside function. How to cope with the last inverse transformation which obviously comes down to inverting of the expression $$ \frac{exp\{(-\frac{k}{2}-\frac{\sqrt{k^2+4s}}{2})(x-y)\} }{\sqrt{k^2+4 s}} $$ with respect to s?