Order and Degree of Ordinary Differential Equation What is the degree and order of the following O.D.E.?
$$\frac{d^2y}{dx^2} + \cos (\frac{dy}{dx}) = 0$$
My view:- Since highest power derivative is $\frac{d^2y}{dx^2}$, degree is 1 and order is 2.
Confusion: this is not in polynomial form, so can we define degree and order?
 A: As given $$\frac{d^2y}{dx^2} + \cos (\frac{dy}{dx}) = 0$$ is a second order nonlinear differential equation.
We can reduce the order setting $p=\frac{dy}{dx}$, which makes $$\frac{dp}{dx} + \cos (p)=0\implies \frac{dx}{dp}+\sec(p)=0$$ which makes (using the tangent half-angle substitution) $$x+c_1=-2 \tanh ^{-1}\left(\tan \left(\frac{p}{2}\right)\right)$$ $$p=-2 \tan ^{-1}\left(\tanh \left(\frac{x}{2}+c_2\right)\right)$$ from which we start with a small nightmare.
A: This is a second-order nonlinear ordinary differential equation, in order to solve it, let $\text{y}'\left(x\right)=\text{r}\left(x\right)$ (as @ClaudeLeibovici mentioned):
$$\text{y}''\left(x\right)+\cos\left(\text{y}'\left(x\right)\right)=0\space\Longleftrightarrow\space\text{r}'\left(x\right)+\cos\left(\text{r}\left(x\right)\right)=0$$
Now:
$$\int\text{r}'\left(x\right)\sec\left(\text{r}\left(x\right)\right)\space\text{d}x=\int-1\space\text{d}x$$
Use:


*

*Substitute $\text{u}=\text{r}\left(x\right)$ and $\text{d}\text{u}=\text{r}'\left(x\right)\space\text{d}x$:
$$\int\text{r}'\left(x\right)\sec\left(\text{r}\left(x\right)\right)\space\text{d}x=\int\sec\left(\text{u}\right)\space\text{d}\text{u}$$

*Substitute $\text{s}=\tan\left(\text{u}\right)+\sec\left(\text{u}\right)$ and $\text{d}\text{s}=\left(\sec^2\left(\text{u}\right)+\tan\left(\text{u}\right)\sec\left(\text{u}\right)\right)\space\text{d}\text{u}$:
$$\int\sec\left(\text{u}\right)\space\text{d}\text{u}=\int\frac{1}{\text{s}}\space\text{d}\text{s}=\ln\left|\text{s}\right|+\text{C}=\ln\left|\tan\left(\text{u}\right)+\sec\left(\text{u}\right)\right|+\text{C}$$

*$$\int-1\space\text{d}x=-\int1\space\text{d}x=-x+\text{C}=\text{C}-x$$


So, we get:
$$\ln\left|\tan\left(\text{r}\left(x\right)\right)+\sec\left(\text{r}\left(x\right)\right)\right|=\text{C}-x$$
Set $\text{y}'\left(x\right)=\text{r}\left(x\right)$ back:
$$\ln\left|\tan\left(\text{y}'\left(x\right)\right)+\sec\left(\text{y}'\left(x\right)\right)\right|=\text{C}-x\space\Longleftrightarrow\space\left|\tan\left(\text{y}'\left(x\right)\right)+\sec\left(\text{y}'\left(x\right)\right)\right|=\frac{\text{C}}{e^x}$$
