# Order and degree of $\exp({{d^3y}/{dx^3}} )- x \frac{d^2y}{dx^2} + y=0$?

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.

• But I've been taught that all the differential coefficients must be free from radical signs and should not appear as an argument of some other function. Can you please confirm whether the above should be true for all differential coefficients or only the coefficient of highest order ? – So Lo Jun 24 '18 at 11:21
• I think that degree should be not defined since the ODE cannot be written as a polynomial in differential coefficients – So Lo Jun 24 '18 at 11:32
• I investigated cause i wasn't sure of my answer and found these links, and you're right, the order it's not defined quora.com/… math.stackexchange.com/q/1117694/546265 – MR ASSASSINS117 Jun 24 '18 at 11:36
• thank you for helping – So Lo Jun 24 '18 at 12:02