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$$y'-2y=e^{ax}\text{ , } y(0)=0$$ Is the solution 1. continuous for every $a$? 2. differentiable for every $a$ ?

I solve the ODE and got $y=\frac{1}{a-2}(e^{ax}-e^{2x})$ so it obvious that at $a=2$ it is not continuous but maybe differentiable?

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    it must be $$a\ne 2$$ the belongs not to the range of defintion thus it makes no sence to ask if the function is continuous for $a=2$ – 2017-01-18
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    Do you mean continuous/differentiable as a function of $x$ for fixed $a$, as a function of $a$ for fixed $x$, or as a function of $a$ and $x$ jointly? In all cases the answer is yes: in fact the solution is analytic jointly in $x$ and $a$. – 2017-01-18
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    The singularity at $a=2$ is removable. – 2017-01-18
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    @RobertIsrael Yes, but yet it is not continuous, we need to "fix" it – 2017-01-18
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    The actual solution is continuous. The formula you gave must be "fixed" by defining it for $a=2$, and then it is continuous. – 2017-01-18

2 Answers 2

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If $a\ne 2$, then of course, you have a solution you gave above. If $a=2$, then we have completely different solution: $y=xe^{2x}$.

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Explicitly, the solution can be expanded in a joint Maclaurin series

$$ y(x,a) = \sum_{n=1}^\infty \sum_{k=0}^{n-1} \frac{2^{n-1-k}}{n!} a^k x^n $$

which converges absolutely for all $a$ and $x$, uniformly on compact sets, so is analytic in $a$ and $x$.

More generally, there are theorems on the dependence of solutions of differential equations on a parameter that apply. See e.g. Birkhoff and Rota, "Ordinary Differential Equations" (I don't have it with me at the moment, so I can't quote these theorems).