Order and degree of $\exp({{d^3y}/{dx^3}} )- x \frac{d^2y}{dx^2} + y=0$?
The order will be 3 since it involves $\frac{d^3y}{dx^3}$
For degree to be defined, the differential coefficients must have natural powers, not occur as an argument to some function.
Hence, the degree will not be defined since the differential coefficients occur as an argument of exponential function (or the log function in case we try to take the log)
Is this correct?
$$e^{y'''}=xy''-y\qquad \rightarrow \qquad y'''=\ln(xy''-y)$$
The ODE order is defined as $\ y^{(n)}\ $ where $\ n\ $ is the derivative and $\ n\ge0\ $ it must be a natural number.
And the degree is defined as $\ (y^{(n)})^{k}\ $ where $\ k\ $ is the exponent and also $\ k>0\ $ it must be a natural number
$\ y'''\ $ is the highest derivative, $\ n=3\ $ and $\ k=1\ $
because the order $\ \ln(y')\ $ isn't defined, as well $\ \ln(xy''-y)\ $ too.