Solution of $\frac{d^2y}{dx^2} - \frac{H(x) y}{b} = H(-x)$ Does the equation

$$\frac{d^2y}{dx^2} - \frac{H(x)}{b} y = c H(x)$$

have a solution where $H(x)$ is the Heaviside step function and $b$ and $c$ are constant?



Update: What about the second step function be $H(-x)$:
$$\frac{d^2y}{dx^2} - \frac{H(x)}{b} y = c H(-x)$$
For $x&lt;0$, $y''=0$, i.e. $$y=C_1x+C_0.$$

For $x\ge0$, $y''-\frac1by=1$.

Assuming $b&gt;0$, the solution of the homogenous equation is
$$y=A\exp(\frac x{\sqrt b})+B\exp(-\frac x{\sqrt b}),$$
and a particular solution is
$$y=-b.$$
Hence the general solution,
$$y=A\exp(\frac x{\sqrt b})+B\exp(-\frac x{\sqrt b})-b.$$

You will ensure continuity of the function by equating the function and first derivative at $x=0$:
$$C_0=A+B-b\\C_1=\frac A{\sqrt b}-\frac B{\sqrt b}.$$

In particular, starting from the steady state, $A=B=\frac b2$, and
$$y=b\left(\cosh(\frac x{\sqrt b})-1\right)H(x).$$
I am not sure wether this answer is correct or not, but as far as the step function is only a sign, I brought it out of the integration.

Please note that, for finding the constants, you need the boundary values and compare the value of the function at the point $x=0$.

$$\begin{array}{l}\frac{{{d^2}y}}{{d{x^2}}} - \frac{{H(x)y}}{b} = H(x)\\\frac{{{d^2}y}}{{d{x^2}}} = \frac{{H(x)y}}{b} + H(x)\\\frac{{{d^2}y}}{{d{x^2}}} = \left( {\frac{y}{b} + 1} \right)H(x)\\\frac{{{d^2}y}}{{\left( {\frac{y}{b} + 1} \right)}} = H(x){d^2}x\\\int {\frac{b}{{y + b}}{d^2}y}  = H(x)\int {{d^2}x} \\\int {\left\{ {bLn\left( {y + b} \right) + {c_1}} \right\}dy}  = H(x)\int {x + {c_3}dx} \\b\left\{ {\left( {y + b} \right)Ln\left( {y + b} \right) - \left( {y + b} \right)} \right\} + {c_1}y + {c_2} = H(x)\left\{ {\frac{1}{2}{x^2} + {c_3}x + {c_4}} \right\}\end{array}$$