The solution to a non-homogeneous differential equation is given by $$ u_r = (u_r)_{homo} + (u_r)_{part}$$ Where $(u_r)_{homo}$ is the homogeneous solution and $(u_r)_{part}$ is the particular solution. I'm trying to find the solution to a differential equation of the form: $$\frac{d^2u_t}{dt^2} + \frac{1}{t} \frac{du_t}{dt} - \frac{u_t}{t^2} = g(t) \tag{1}$$ However; I am only familiar with solving differential equations of the form: $$\frac{d^2u_t}{dt^2}+A\frac{du_t}{dt}+Bu = g(t) \tag{2}$$ Where A and B are real numbers (e.g. A=2 and B=4) and are not functions of t. I know the homogeneous solution to $\text{(1)}$ but am having difficulty finding its particular solution. The homogeneous spolution is shown below: $$(u_r)_{homo} = C_1r + \frac{C_2}{r}$$ Can someone help me with solving the differential equation $\text{(1)}$ or at least tell me the name of that form of differential equation so I could search for a guide?

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    $\begingroup$ In Mathematica, you can try DSolve[u''[t] + u'[t]/t - u[t]/t^2 == a Sin[t], u[t], t] which returns a reasonably nice answer. It gives an answer for arbitrary g(t) (instead of a Sin[t]). $\endgroup$
    – bill s
    Apr 9, 2016 at 2:15
  • $\begingroup$ Make substitution $s=\ln t$ (or $s=\ln r$ if you prefer to call your variable $r$. There is a small confusion on that point in the question). $\endgroup$
    – mickep
    Apr 9, 2016 at 15:29
  • $\begingroup$ Whoops, I originally wrote everything in terms of $r$ and $u_r$ but realized it is more common notation to use $t$ and $u_t$ so I started changing to that but I forgot quite a few... $\endgroup$ Apr 9, 2016 at 17:42

1 Answer 1


It's called Euler's differential equation and may be converted to a linear differential equation with constant coefficients by the substitution $v=\ln t$ so $t=e^v$ and $$\begin{align}\frac{du_t}{dv} & =\frac{du_t}{dv}\frac{dv}{dt}=\frac{1}{t}\frac{du_t}{dv}\\ \frac{d^2u_t}{dt^2} & =-\frac{1}{t^2}\frac{du_t}{dv}+\frac{1}{t^2}\frac{d^2u_t}{dv^2}\end{align}$$ Then the differential equation will read $$\frac{d^2u_t}{dv^2}-u_t=e^{2v}g(e^v)$$


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