<s>[INST] 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 $H(-x)$:
$$\frac{d^2y}{dx^2} - \frac{H(x)}{b} y = c H(-x)$$
 [/INST] 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 $x=0$.
$$\begin{array}{l}\frac{{{d^2}y}}{{d{x^2}}} - \frac{{H(x)y}}{b} = H(x)\\\frac{{{d^2}y}}{{d{x^2}}} = \frac{{H(x)y}}{b} + H(x)\\\frac{{{d^2}y}}{{d{x^2}}} = \left( {\frac{y}{b} + 1} \right)H(x)\\\frac{{{d^2}y}}{{\left( {\frac{y}{b} + 1} \right)}} = H(x){d^2}x\\\int {\frac{b}{{y + b}}{d^2}y}  = H(x)\int {{d^2}x} \\\int {\left\{ {bLn\left( {y + b} \right) + {c_1}} \right\}dy}  = H(x)\int {x + {c_3}dx} \\b\left\{ {\left( {y + b} \right)Ln\left( {y + b} \right) - \left( {y + b} \right)} \right\} + {c_1}y + {c_2} = H(x)\left\{ {\frac{1}{2}{x^2} + {c_3}x + {c_4}} \right\}\end{array}$$
</s>