$$ m\frac {d^{2}y}{d^{2}x} = 1 $$

homogenous linear equation for this ode is

$$ m\frac {d^{2}y}{d^{2}x} = 0 $$

trial solution is $Ae^{kx}$ but clearly in this case $Bx+C$ is a trial solution that works. What is the logic behind having 2 trial solutions only in this case? For every other problem I only guess the exponential solution.

  • $\begingroup$ Well, clearly a polynomial of degree $1$ has second derivative equal to zero... $\endgroup$
    – Math1000
    Commented Apr 27, 2018 at 18:26

2 Answers 2


Well, the trial solution is still $e^{kx}$ for some $k$ we should determine. Plugging it in, we get:

$$m\frac{d^2y}{dx^2}=0 \implies mk^2e^{kx}=0 \implies k^2=0 \implies k=0,0$$

So, the answer should be of the form $y(x)=Ae^{0x}+Be^{0x}=C$, for some constant $C$.

However, a second order linear ODE should have two linearly independent solutions. As such, it turns out that whenever we have a repeated solution like the above, we can get another solution by multiplying the current one by $x$.

Indeed, we find that $Dx$ is also a solution.

Hence, the general solution is $C+Dx$


The solution $Bx+C$ is a limit for the combination of the solutions $Ae^{\pm \omega\,x}$, when $\omega \,\to \,0$.

In fact, when you consider the general linear 2nd order ODE $$ m{{d^2 y} \over {d^2 t}} + r{{dy} \over {dt}} + ky = 0 $$ and write the general solution to it, in case of an under-damped system, as $$ \eqalign{ & f(t) = c_{\,1} e^{\,\rho \,t + i\,\omega \,t} + c_{\,2} e^{\,\rho \,t - i\,\omega \,t} = \left( {c_{\,1} e^{\,i\,\omega \,t} + c_{\,2} e^{\, - i\,\omega \,t} } \right)e^{\,\rho \,t} = \cr & = \left( {a\cos \,\left( {\omega \,t} \right) + b\sin \,\left( {\omega \,t} \right)} \right)e^{\,\rho \,t} \cr} $$ where $$ \omega = \sqrt {k/m - \left( {r/\left( {2m} \right)} \right)^2 } \quad \quad \rho = r/\left( {2m} \right) $$

and impose the initial conditions, for instance for $f(0)$ and $f'(0)$, you get $$ \left\{ \matrix{ f(0) = a \hfill \cr f'(0) = \,\,b\omega + \rho a\quad \Rightarrow \quad b = \;{1 \over \omega }\left( {f'(0) - \,\,\rho f(0)} \right) \hfill \cr} \right. $$ so $$ f(t) = \left( {f(0)\cos \,\left( {\omega \,t} \right) + {1 \over \omega }\left( {f'(0) - \,\,\rho f(0)} \right)\sin \,\left( {\omega \,t} \right)} \right)e^{\,\rho \,t} $$

Now, if the damping approaches the critical value, that is $\omega \to 0$, then $$ \bbox[lightyellow] { \mathop {\lim }\limits_{\omega \, \to \,0} f(t) = \left( {f(0) + \left( {f'(0) - \,\,\rho f(0)} \right)t} \right)e^{\,\rho \,t} }$$ and when also $\rho$ (that is $r$) approaches $0$ you get the $Bt+C$.

Same if considering an over-damped system as explained in this related post.


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