How to solve $\mathrm{d}^n y/\mathrm{d}x^n = (\mathrm{d}y/\mathrm{d}x)^n$ or $(\mathrm{e}^D) y = \mathrm{e}^{(Dy)}$

127 Views Asked by At

Is there a general analytical solution for $$ \frac{\mathrm{d}^ny}{\mathrm{d}x^n} = \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^n,\qquad n>2? $$

or equivalently $(\mathrm{e}^D) y = \mathrm{e}^{(Dy)}$ where $\mathrm{D} = \frac{\mathrm{d}}{\mathrm{dx}}$ and $\mathrm{e}^D$ is the exponentiatin operator.

For $n=2$, the solution is $y(x) = c_2 - \log(c_1 + x)$.