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Problem: Solve the initial value problem $y’=3y; y(0)=a>0$.

My solution: Exploiting variable separable method, $${dy\over y}=3\, dx$$ $$\implies\ln |y|=3x+C$$ where $C$ is the constant of integration. Now using $y(0)=a$ and $a>0$, $\ln a=C$. $$\therefore\ln |y|=3x+\ln a$$ $$\implies |y|=e^{3x+\ln a}$$ $$\implies y=\pm ae^{3x}$$

But clearly $y=-ae^{3x}$ isn’t a solution for the given problem.

So what went wrong?

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There is no error, as $-ae^{3x}$ is not a solution, violating the initial condition, and $+ae^{3x}$ is a solution, you have found the unique solution.

It is always possible that some non-equivalent step increases the number of intermediate solution candidates, so that you have to compare back with the original problem to select the true solutions.


You could also have done the preliminary analysis of the ODE in that $y=0$ is a constant, stationary solution of the ODE, so that by uniqueness no other solution can change its sign (as a sign change would imply the crossing of two solutions). Then because of $y(0)=a>0$, the solution of the IVP is positive, so that you can resolve the absolute value in $\ln|y|$ immediately to $\ln y$, removing the ambiguity.


Or staying with your solution method, observe that by dividing by $y$ the result is only valid as long as $y\ne 0$ to avoid division by zero. By the initial condition this means the obtained solution is only valid where $y>0$, again giving $|y|=y$. In the finished solution you can then find that the solution function has no zero crossings, thus is valid everywhere.

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  • $\begingroup$ Can you please explain why $y=0$ being a solution for the ODE implies that no solution can change sign. $\endgroup$ – Atom Jul 20 at 13:58
  • $\begingroup$ Because the right side is smooth, all solutions are (locally) unique. If some solution is zero somewhere, it is already the zero solution everywhere. $\endgroup$ – LutzL Jul 20 at 14:17
  • $\begingroup$ Not getting your second statement. And how does it lead to what I’m asking? Please help. $\endgroup$ – Atom Jul 20 at 14:34
  • $\begingroup$ Assume you have some arbitrary solution where you only know that it has a zero somewhere. Then by uniqueness you automatically know that it is already the zero solution. Changing sign requires by IVT that there is a zero somewhere, which is thus impossible. $\endgroup$ – LutzL Jul 20 at 14:39
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    $\begingroup$ No, what I'm saying is that a scalar ODE of the form $y'=f(y)=yg(y)$ has the zero function as solution and thus any IVP with an non-zero initial value can have no zeros. More generally, in an autonomous scalar ODE $y'=f(y)$ any solution is bounded by the stationary solutions at the roots of $f(y)=0$, if an initial value is between two such roots, then the full solution stays between these roots. $\endgroup$ – LutzL Jul 20 at 14:53
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The implications are only flowing forward, not backwards. You know, from your argument, $$((y' = 3y) \land (y(0) = a > 0)) \implies (y = ae^{3x}) \lor (y = -ae^{3x}).$$ You don't know the converse: $$(y = ae^{3x}) \lor (y = -ae^{3x}) \implies ((y' = 3y) \land (y(0) = a > 0)).$$ In fact, the converse is false, as you noted. If $y = -ae^{3x}$, then the right side is not satisfied.

This is a common thing in many equation-solving methods: an infinite set of potential solutions is reduced to a small finite set of possible solutions, but none of the potential solutions are actually proven by the method. There may even be no solutions! It is important to verify the potential solutions you obtain from such a solving method, otherwise you might settle on erroneous possible solutions.

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    $\begingroup$ I remember once trying to prove the existence of a solution of an ODE. A friend (and a more senior mathematics student) plugged it into Matlab to obtain an approximate solution, and argued that this proved the existence of a solution. This was an example of a similar fallacy. Most numerical procedures for approximating solutions to DEs tend to blindly assume that solutions exist. Just because something that looked like a suitable curve popped out of the algorithm, doesn't mean that there was a solution that was being approximated. $\endgroup$ – Theo Bendit Jul 20 at 13:44

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