Implicit solution to ODE or just error?

Question

Consider the ODE $$y''+y'=2y$$ For what $a$ is $e^{ay}$ a solution?

I'm wondering if the question was supposed to contain $e^{ax}$ instead of $e^{ay}$, since if it is $e^{ay}$ then substitution gives

$$y'=ay'e^{ay}, y''=ay''e^{ay}+(ay')^2e^{ay}$$ so that the ODE becomes

$$ay''e^{ay}+(ay')^2e^{ay}+ay'e^{ay}=2e^{ay}$$ or

$$e^{ay}(ay''+(ay')^2+ay'-2)=0$$

and now we have

$$ay''+(ay')^2+ay'=2$$

So we end up with a more complicated equation. Please let me know what you think.

Answer 1

Assuming $y = y(x)$, first note that the general solution is $y = c_1 e^{-2x} + c_2 e^{x}$.

It looks to me like you are trying to reverse engineer the technique of guessing an ansatz but for a function of $y$ instead of $x$. The procedure works by supposing the solution is in the form $y = c_1 e^{\lambda x} + c_2 e^{\mu x}$ and you end up with the familiar quadratic that determines the appearance of $\sin(x), \cos(x), xe^x$ terms etc. It does not really make sense to guess a "solution" $y = f(y(x))$ since all you could hope to achieve is just get a strange relationship about $y(x)$, assuming such a thing can be obtained.

What you are doing is just transforming the ODE, which is definitely useful in other situations (polar coordinates in certain systems works nicely for example), though not so much here since its original form is already nice. The ODE you have should be the same though, just from some guesses by eye $y(x) = k_1 -2 x + k_2 + x$ works and looks like it agrees with the transformation you performed, but note this is a function of $x$ not of $y$.

Finally, suppose we ignored this and there was a nonsensical solution $f(y)$, we already have a G.S above which is clearly a function of $x$, so $f(y)$ is either a constant (put that in the ODE to see it wont work) or not the same, but this contradicts the theorems of existence and uniqueness of a solution.

I hope I have interpreted your question correctly.

Answer 2

For y=f(x) we get $$y''+y'=2y\implies y=C_1e^{-2x}+C_2e^{x}$$

For $y=e^{ay}$, you do not need a differential equation. $$ y=e^{ay} \implies ln y=ay\implies ln y /y =a$$

This equation has one solution for $a\le 0$,or $a=e$, two solutions for $0<a<e$, and no solutions otherwise.

These solutions are all constant solutions and none of them satisfy the given differential equation.