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I was wondering if there is an analytical solution to the following homogeneous linear differential equation $$\dfrac {dM} {dt}=\dfrac {M} {\alpha \left( t\right) }e^{\beta\left( t\right) t}$$ which could obviously be rewritten as $$\int dM =\int \dfrac {M} {\alpha \left( t\right) }e^{\beta\left( t\right) t}dt$$ Where both $\alpha$ and $\beta$ are unknown functions of $t$. I am some what unsure of integrating $\alpha$ and $\beta$ function parts. Valid ranges for $M={0 to \inf}$ and $t={0 to \inf}$. Any help would be much appreciated.

Edit: Is the following evaluatable $\int \dfrac {e^{\beta\left( t\right) t}} {\alpha \left( t\right) }dt$ ?

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Your equation is "separable": $$\frac{dM}{M} = \frac{e^{\beta(t) t}}{\alpha(t)}\;dt$$ – GEdgar Mar 21 '12 at 16:42
Since you don't know anything useful about $\alpha (t),\ \beta (t)$, you cannot evaluate integrals explicitly. So, if your question is Does $M(t)$ have a simple elementary form?, then it is totally unanswerable. On the other hand, if your question is Does my ODE have a solution?, then the answer is Yes, it does (under mild assumptions on $\alpha (t),\ \beta (t)$). – Pacciu Mar 21 '12 at 17:02
@Pacciu thanks so much for that clarification. One more question i know this must sound extremely dumb of me , but do u think it would be ok if i assumed $\alpha$ and $\beta$ to be constants did the integration and in the solution promoted $\alpha$ and $\beta$ to functions of t again. Would that be completely wrong or could i justify a this as an estimate of my problem doing this ? – Comic Book Guy Mar 21 '12 at 17:08
It would be completly wrong, of course. The question is: what do you know about $\alpha$ and $\beta$? Are they just arbitrary functions, or do they have some good properties (e.g., countinuity, boundedness,...)? – Pacciu Mar 21 '12 at 17:09
@Hardy: I was thinking about continuity. Please read my answer below. ;-) – Pacciu Mar 21 '12 at 18:21
up vote 1 down vote accepted

If you don't know anything useful about $\alpha (t)$ and $\beta (t)$, you can't evaluate the general integral of your ODE. Neverthless, you can gather some properties of your ODE's solutions (under mild assumptions on $\alpha,\ \beta$) with the aid of a qualitative study of the maximal solutions.

You have the following ODE: $$M^\prime(t) = \frac{M(t)}{\alpha (t)}\ e^{t\ \beta (t)}\; ;$$ assume that $\alpha ,\beta :\mathbb{R}\to \mathbb{R}$ are continuous.

The RH side of your ODE, i.e.: $$f(t,M;\alpha ,\beta):=\frac{M}{\alpha (t)}\ e^{t\ \beta (t)}$$ is well-defined as a function of $(t,M)$ in $\Omega \times \mathbb{R}$, where $\Omega:=\{t\in \mathbb{R}:\ \alpha (t)\neq 0\}\subseteq \mathbb{R}$ is assumed to be a nonempty open set (because, if it were empty, then $\alpha (t)=0$ for every $t$ and there won't be any possible meaning for the RH side).

Your $f(\cdot ,\cdot ; \alpha ,\beta)$ is a continuous function in $\Omega \times \mathbb{R}$ , hence your ODE has always at least a local solution (by Peano's Existence Theorem) satisfying $M(t_0)=M_0$ for any $(t_0,M_0)\in \Omega \times \mathbb{R}$; moreover such a local solution is in fact a $C^1$ function in a neighbourhood of $t_0$ (and, in general, if $\alpha (t)$ and $\beta (t)$ are of class $C^k$, then $M(t)$ is of class $C^{k+1}$).

On the other hand, your $f(t,M;\alpha ,\beta)$ is a linear function of $M$, meaning that for each $t\in \Omega$ the map $f(t,\cdot;\alpha ,\beta)$ is linear. Therefore $f(t,\cdot ;\alpha ,\beta)$ is locally Lipschitz and the Local Uniqueness Theorem applies: thus there is only one local solution of your ODE matching the initial condition $M(t_0)=M_0$ (for each $(t_0,M_0)\in \Omega \times \mathbb{R}$).

Now, chose $(t_0,M_0)\in \Omega \times \mathbb{R}$ and consider the Cauchy's problem: $$\tag{CP} \begin{cases} M^\prime (t) = \frac{e^{t\ \beta (t)}}{\alpha (t)}\ M(t)\\ M(t_0)=M_0\; . \end{cases}$$ It can be proved that the unique local solution of (CP) can be extended untill its graph "touches" $\partial (\Omega \times \mathbb{R})$: this extension gives you the socalled maximal solution of (CP).

If $M_0=0$, since $f(t,0;\alpha, \beta)=0$, the function $M^*(t)=0$ is a stationary solution of (CP) and it is defined in the largest interval $I_{t_0}\subseteq \Omega$ which contains $t_0$.

Since $f(t,M ;\alpha ,\beta)> 0$ if and only if $(t,M)\in (\Omega^+ \times ]0,\infty[ )\cup (\Omega^- \times ]-\infty, 0[)$, where $\Omega^\pm :=\{ t\in \mathbb{R}:\ \alpha (t) \gtrless 0\}$, it is clear that your maximal solution will be either increasing or decreasing depending on whether $(t_0,M_0)\in (\Omega^+ \times ]0,\infty[ )\cup (\Omega^- \times ]-\infty, 0[)$ or $(\Omega^- \times ]0,\infty[ )\cup (\Omega^+ \times ]-\infty, 0[)$.

If one of $\Omega^+$ or $\Omega^-$ contains a neighbourhood of $+\infty$, say $]T,+\infty[$, then any maximal solution of (CP) with $t_0\in ]T,+\infty[$ can be extended to the whole $]T,+\infty[$ (because $f(t,\cdot; \alpha ,\beta)$ is linear); the same if one of of $\Omega^+$ or $\Omega^-$ contains an interval of the type $]-\infty, T[$.

Obviously, the maximal solution of (CP) is given by: $$M(t)=M_0\ \exp \left(\int_{t_0}^t \frac{e^{\tau\ \beta (\tau)}}{\alpha (\tau)}\ \text{d} \tau \right)$$ and you could gather other informations on $M$ from this formula (hopefully!).

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wow, that was mind blowing. I am not sure if understood all of it in a quick read but i can Google and decipher the terms you mentioned along with reading up on those theorems. Thank you so much you are a life saver. – Comic Book Guy Mar 21 '12 at 19:04

hint: $\frac{d}{dt} \log(M)=?$

edit: Integrate both sides from $0$ to $t$ to obtain $$\log M(t)-\log (M(0))=\int_0^t \frac{e^{\beta(s)s}}{\alpha(s)}ds.$$ Solve for $M(t)$

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Hey buddy so you first seperated them integrated the M side to get log M and then brought back the dt in RHS numerator to LHS denominator. Is that right ? although with that said $\int \dfrac {dM} {M}=\int \dfrac {e^{B\left( t\right) t}} {\alpha \left( t\right) }dt$ which gives $\dfrac {d} {dt}\left( \log M+c\right) =\dfrac {e^{B\left( t\right) t}} {\alpha \left( t\right) }$ I am not sure how to solve this, any further clues would be much appreciated. – Comic Book Guy Mar 21 '12 at 16:51

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