Show that $y=e^{e^{cx}}$ is a solution of the differential equation $\frac{d^2y}{dx^2} =c^2 \cdot y \cdot \ln(y) (1+\ln(y))$

Question

Question: Show that $y=e^{e^{cx}}$ is a solution of the differential equation $$\frac{d^2y}{dx^2} =c^2 \cdot y \cdot \ln(y) (1+\ln(y))$$

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I know there are a lot of ways of solving this and I was trying this way and I don't seem to get the answer however I get close to it

What I have done:

$$y=e^{e^{cx}}$$

$$ ⇔ \ln(y) = e^{cx} $$

$$ ⇔ \ln(\ln(y)) = cx $$

Derivative of both sides

$$ \frac{1}{y \ln(y)} \frac{dy}{dx} = c$$

$$ \frac{dy}{dx} = c (y \ln(y)) $$

Second derivative now

$$ \frac{d^2y}{dx^2} = c (\ln(y)+y \cdot \frac{\frac{dy}{dx}}{y}) $$

$$ \frac{d^2y}{dx^2} = c (\ln(y)+ \frac{dy}{dx}) $$

But $ \frac{dy}{dx} = c (y \ln(y)) $

$$ \frac{d^2y}{dx^2} = c (\ln(y)+ c (y \ln(y))) $$

$$ \frac{d^2y}{dx^2} = c \ln(y)+ c^2y \ln(y) $$

$$ \frac{d^2y}{dx^2} = c^2 y \ln(y)(\sqrt{c} \ln(y) + 1) $$

Which is close but not correct where exactly did I go wrong?

Answer 1

The problem is when you calculated the second derivative and left out a $\frac{dy}{dx}$ term: $$\frac{d^2 y}{dx^2} = c \Big(\ln(y) \frac{dy}{dx} + \frac{dy}{dx} \Big).$$