I have the following differential equation: $$\frac{d^{2}y}{dx^{2}} + \frac{1}{b^{2}} y = -\frac{\pi}{b} J(1, x).$$ $J$ is the Bessel function of first type. In order to be able to solve it exactly (at least according to Mathematica), I need to change the variables so that the two terms on the LHS have equal coefficients. So, I try $x = b t$, and get $$\frac{1}{b^{2}}\frac{d^{2}y}{dt^{2}} + \frac{1}{b^{2}} y = -\frac{\pi}{b} J(1, bt)$$ Is this correct? Is this how I change variables in a differential equation, especially in the inhomogeneous term? When I find a solution, what variable is it in exactly? How do I recover the solution in terms of $x$?
Thanks.
If you set $x=b t$ then $dx=b\cdot dt$ and you go from $$ \frac{d^2 y}{dx^2}+\frac{1}{b^2}\cdot \frac{dy}{dx} = -\frac{\pi}{b}\cdot J_1(x) \tag{1}$$ to $$ \frac{1}{b^2}\cdot\frac{d^2 y}{dt^2}+\frac{1}{b^\color{red}{3}}\cdot \frac{dy}{dt} = -\frac{\pi}{b}\cdot J_1(bt). $$ $(1)$ can be solved through Frobenius power series method, or through the Laplace transform. By applying $\mathcal{L}$ to both sides of $(1)$ and setting $g(s)=\left(\mathcal{L} y\right)(s)$ we get
$$ -\left(s+\frac{1}{b^s}\right) y(0)-s y'(0) + \left(s^2+\frac{s}{b^2}\right) g(s) = -\frac{\pi}{b\sqrt{1+s^2}(s+\sqrt{1+s^2})}\tag{2} $$ hence $g(s)$ has an explicit closed form in terms of $y(0),y'(0)$ and $y$ can be recovered from $\mathcal{L}^{-1} g$.
The edited version can be tackled in the same way: the general solution to $$ y''+\frac{1}{b^2}y=-\frac{\pi}{b}J_1(x) $$ is given by $$ C_1 \cos\frac{x}{b}+C_2\sin\frac{x}{b}+\pi\int_{x_0}^{x}J_1(z)\sin\frac{z-x}{b}\,dz.\tag{3} $$