# Particular integral for $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=\ln x$

I would like to know how to find a particular integral for $$a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=\ln x$$ where $$a,b,c$$ are constants. So far, the only functions I've come across for $$f(x)$$ in $$a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=f(x)$$ are $$f(x)=x^ne^{kx}$$ $$f(x)=A\sin\alpha x+B\cos\alpha x$$ $$f(x)=\text{polynomial in x with positive integer exponents}$$ and the particular integrals I've been taught to use respectively are $$x^ne^{kx}+x^{n-1}e^{kx}+\cdots+e^{kx}$$ $$y=C\sin\alpha x+D\cos\alpha x$$ $$y=\text{polynomial in x with positive integer exponents of the same degree as f(x)}$$ unless complications arise when finding the complimentary function.

However, how do I find the particular integral for when $$f(x)=\ln x$$?

• You will need the Wronskian. – John Wayland Bales Jan 16 at 22:17
• Unfortunately, $\ln x$ is not one of the functions where the method of undetermined coefficients works. So some other method is needed. ODE textbooks should discuss variation of parameters. – GEdgar Jan 16 at 22:59

For a linear, second-order, constant-coefficient ODE, with arbitrary function $$h(x)$$, \begin{align} a\frac{\mathrm d^2y}{\mathrm dx^2}+b\frac{\mathrm dy}{\mathrm dx}+cy=h(x) \end{align} making the substitution \begin{align} z(x)=\delta y(x)+\frac{\mathrm dy}{\mathrm dx}, \end{align} \begin{align} \delta =\left(\frac{b\pm\sqrt{b^2-4ac}}{2a}\right) \end{align} reduces the order of your ODE: \begin{align} a\frac{\mathrm dz}{\mathrm dx}+\frac{c}{\delta}z(x)=h(x). \end{align} I think you can take it from there.