How to find the degree of following differential equation? $$\frac{d^2y}{dx^2}=\sin\left(\frac{dy}{dx}\right) $$
As differential coefficients are not in polynomial function the degree is not defined.
But we can write the this by expanding sine function by Maclaurin series.
$$\frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)-\frac{\left(\frac{dy}{dx}\right)^3}{3!} \cdots $$
Now its a  polynomial. Now the degree must be $1$. But the degree is not defined. 
 A: In something like $\left( \dfrac{dy}{dx} \right)^3$, you don't have a third-order derivative $\dfrac{d^3y}{dx^3}$.  The highest order of derivatives here is $2$, on the left side where you have $\dfrac{d^2y}{dx^2}$.  So you could say it's a second-order differential equation.
However, I would write $$v=\frac{dy}{dx} \tag 1$$ so that $\dfrac{dv}{dx} = \dfrac{d^2y}{dx^2}$ and then you have
$$
\frac{dv}{dx} = \sin v,
$$
which is a first-order differential equation.  Then find $y$ as a function of $x$ by antidifferentiating both sides of $(1)$.
A: 
Now its a polynomial

No it isn't. You still have an infinite series, which is not a polynomial.
A: Because i found this an interesting problem,
 i want to give an exact solution of the differential equation
(1) Reduction of order $y'(t)=p(t)$
$$
p'(t)=\sin(p(t))
$$ 
(2) Reduce it to a separable form: $p(t)=\arcsin(q(t))\rightarrow p'(t)=\frac{q'(t)}{\sqrt{1-q^2(t)}}$
$$
q'(t)=q(t)\sqrt{1-q^2(t)}\rightarrow\\
dt=\frac{dq}{q\sqrt{1-q^2}}
$$
(3) perform the integration (they are standard)
$$
t-t_0=\log\left(\frac{q}{1+\sqrt{1-q^2}}\right)
$$
(4) backsubsitution  $q=\sin(p)$ and algebra 
$$
t-t_0=\log\left(\tan\left(\frac{p}{2}\right)\right)\rightarrow\\
p(t)=2\arctan\left(e^{t-t_0}\right)
$$
Please note the close connection to the soliton solutions of the Sine Gordon equation
(5) To obtain the finalresult we have to integrate $p(t)$ once more.
$$
y(t)=\int dt p(t)=\\
i \left(\text{Li}_2\left(-i e^{t-t_0}\right)-\text{Li}_2\left(i e^{t-t_0}\right)\right)+t_1=2 \text{Ti}_2(e^{t-t_0})
$$
which can be checked by direct differentiation (or integration by parts).
Please note  that $\text{Li}_2\left(x\right)\quad, \text{Ti}_2(x)$ denote the dilogarithmic function and the inverse tangent integral 
