I have an ODE: $$\ddot v=-kt^{-1}\dot v+kt^{-2}v$$ I want to solve it by reducing it to a first order ODE, by defining $u(t)=t^{m}v(t)$ If I rewrite the ODE in terms of $u$, this gives me: $$\ddot u =(-k+2m)t^{-1}\dot u +(km+k-1-m^2)t^{-2}u$$ Setting the coefficient of $u$ to zero gives $2m=k+\sqrt{k^2-4+4k}$, which I will write as $k+\alpha$.

This gives me the single order ODE $$\ddot u=\alpha t^{-1}\dot u$$ Solving it gives $$\dot u=2c\alpha t$$ $$u(t)=c\alpha t^2$$ Plugging this back into $v$, gives $$v(t)=c\alpha t^{2-m}$$ $$\dot v = (2-m)c\alpha t^{1-m}$$ $$\ddot v = (1-m)(2-m)c\alpha t ^{-m}$$

These equations are supposed to hold for any $k$. If we plug this into the original ODE, however, this reduces to the equation $$3k=4+\sqrt{k^2-4+4k}$$

Which obviously does not hold for all $k$.

What am I doing wrong?

  • 1
    $\begingroup$ Hint: Since this is a Euler-Cauchy type, let $v = t^m $. Find $v''$ and $v'$ and substitute in, solve for $m$ and solve. $\endgroup$
    – Moo
    Mar 30 '17 at 17:39
  • $\begingroup$ most likely your solution changes between (over)damped and oscillatory depending on $k$ $\endgroup$
    – tired
    Mar 30 '17 at 18:00
  • $\begingroup$ Do you need to reduce it to a first order ODE? Or are you fine solving it with any method? $\endgroup$ Mar 30 '17 at 18:06
  • $\begingroup$ I want to learn as much as possible about these things, so I was working on reducing it to a first order ODE. So I'd like to learn about other methods as well, but also I'd like to make sure I can do this method $\endgroup$
    – user56834
    Mar 30 '17 at 18:09
  • $\begingroup$ Please check again all signs and coefficients, the one before $u$ should reduce to $-(m+k)(m-1)$. $\endgroup$ Mar 30 '17 at 18:53

What you should realize is that you don't need to do a change of variables here but recognize that your solution is a Euler-Cauchy problem and therefore can be solved by:

Let $v=t^m$ then subbing the values into your DE you get:

$$ m(m-1)t^{m-2} = -kmt^{m-2} + kt^{m-2} \\ \implies m(m-1) +km -k =0 \\ \implies m=1,m=-k \\$$

therefore giving:

$$ v(t) = At^1 + Bt^{-k} \implies v(t) = At + Bt^{-k}$$

Which does hold $\forall k \in \mathbb{R}$ but you should also note that it does hold $\forall t \in (-\infty,0) \cup (0,+\infty) $ because $t=0$ is a case basis in that if $k>0 $ then $t \neq 0$ but if $k<0$ then $t\in (-\infty,\infty) $

Since you did say you want to learn about the reduction to first order just note then when you do that you typically let u be something like $u= t^m \dot{v} $ so then $\dot{u}$ would be equal to a second derivative of v.


In your first integration, you should get from $$ \ddot u=αt^{-1}\dot u $$ to $$ \ln|\dot u|=α\ln|t|+c $$ or $$ \dot u = C·t^α $$ leading to $$ u=\frac{C}{α+1}·t^{α+1}+D $$


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