Consider the equation \begin{equation}\label{1} f(x,y) := x^2+ye^x-\cos(xy)=0\tag{1} \end{equation} Solve for $y$ in terms of $x$ as a power series. What is the radius of convergence?
My approach to the problem is as follows:
From \ref{1}, we can compute $y(0)$ which is equal to 1.
Taking the derivative of \ref{1} gives $$2x+e^x(y+y^\prime)+\sin(xy)(y+xy')=0 \implies y'(0)=-1$$.
By taking the consecutive derivatives, we are getting $$y''(0)=-2\;\;\; \text{and} \;\;\;y'''(0)=14$$
So, the Taylor Series of $y(x)$ around 0 will be $$y(x)=1-x-x^2+\frac{14}{3!}x^3+\cdots$$
Now, I wanted to ask that
- Is this correct approach?
- If it is correct, then how to determine the radius of convergence here? Otherwise, guide for the right answer.
In principle, your approach is correct. The calculation of the successive derivatives is tedious, though, it's better to solve your differential equation of first order, or rather the equivalent Volterra integral equation $$ y(x)=1+\int^x_0\frac{-2t-y(t)\left(e^t+\sin ty(t)\right)}{e^t+t\,\sin ty(t)}\,dt ,$$ in the algebra of Taylor series, with code like in this sandbox. For the coefficients in $$y(x)=\sum^\infty_{k=0}a_k\,x^k,$$ we obtain $$ \begin{align} a_{0}&=1.0\\ a_{1}&=-1.0\\ a_{2}&=-1.0\\ a_{3}&=2.33333333333333\\ a_{4}&=-1.16666666666667\\ a_{5}&=-3.3\\ a_{6}&=6.625\\ a_{7}&=0.328968253968253\\ a_{8}&=-18.4565476190476\\ a_{9}&=22.7168209876543\\ a_{10}&=24.931390542328\\ a_{11}&=-101.686693973465\\ a_{12}&=57.2597598421383\\ a_{13}&=246.251658434801\\ a_{14}&=-516.944945196425\\ a_{15}&=-105.626413366204\\ a_{16}&=1860.59880450026\\ a_{17}&=-2197.93934423607\\ a_{18}&=-3282.05097142456\\ a_{19}&=12064.9631741049\\ a_{20}&=-5434.85793947368\\\end{align} $$ The numerical values (via root criterion) give an approximation of the radius of convergence, $\approx0.6$. This is confirmed by the values of $f(x,y)$, using the above expansion up to $x^{20}$: $$ \begin{align} f(0.1, y(0.1))&=1.11022302462516e-16\\ f(0.2, y(0.2))&=5.65286706333268e-11\\ f(0.3, y(0.3))&=2.43553489376325e-07\\ f(0.4, y(0.4))&=8.77321908963058e-05\\ f(0.5, y(0.5))&=0.00810985465801761\\ \end{align} $$ The analytical calculation of the radius of convergence is tricky. Obviously, the function $y(x)$ is analytic as long as $e^x+x\,\sin xy\neq0$, and the singularity closest to $x=0$ can be calculated as the (complex!) solution of the system $$ \begin{align} x^2+y\,e^x-\cos xy&= 0\\ e^x+x\,\sin xy&= 0 \end{align} $$ with minimal $|x|$. That should be possible, numerically.