# What is the mistake in this solution?

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?

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.

• Can you please explain why $y=0$ being a solution for the ODE implies that no solution can change sign. – Atom Jul 20 at 13:58
• Because the right side is smooth, all solutions are (locally) unique. If some solution is zero somewhere, it is already the zero solution everywhere. – LutzL Jul 20 at 14:17
• 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. – LutzL Jul 20 at 14:53
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.