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I know how to solve some differential equations assuming that the solution is a Taylor series, but I'm stuck on the following equation. Normally I would equate coefficients, but here I have a cosine term, and it's a cosine of a function of the solution which seems bizarre.

Problem: Find the Taylor series solution to $y^{\prime} = 3\cos(xy+y-4) + y^2 -12$ given that $y(0) = 4$.

I assume that the series is centered at 0 so that $y = \Sigma_{n=0}^\infty c_nx^n$. From the initial value we have that $c_0 = 4$.

Normally I would substitute this solution into the equation. To that end $$ y^{\prime} = \Sigma_{n=1}^\infty nc_nx^{n-1} $$ and $$ xy+y-4 = (4+c_1)x + (c_1+c_2)x^2 + (c_2+c_3)x^3 + \cdots. $$ But then taking the cosine of the above? Does that all simplify somehow?

Ungar Linski
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    You could use Taylor series arithmetic similar to its use in https://math.stackexchange.com/questions/2432062/how-solve-this-nonlinear-trigonometric-differential-equation/2499684#2499684. This is not really a simplification, but it organizes the computation a little bit better. – Lutz Lehmann Feb 16 '22 at 22:13
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    Plugging a series into another series to obtain a third series is possible, basically you end up needing to do a bunch of Cauchy products for each power and then collect the terms of low enough order. This seems hard if you need the whole series, though. – Ian Feb 16 '22 at 22:15
  • @Ian What's a Cauchy product? In this case is it just collecting the coefficients for each power of $x$? – Ungar Linski Feb 16 '22 at 22:19
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    That's all you're really doing, the Cauchy product is just a formula for how the product of two power series actually turns out. https://en.wikipedia.org/wiki/Cauchy_product – Ian Feb 16 '22 at 22:20

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One can make this equation trivial for Taylor calculus using the differential equations for the elementary functions. Then $$ z=xy+y−4\\ u=\cos(z)\\ v=\sin(z)\\ $$ so that the non-linear operations reduce to Taylor series products $$ u'=-vz'\\ v'=uz'\\ y'=3u+y^2-12 $$ For the coefficients of the Taylor expansion this gives recursions $$ z_n=y_{n-1}+y_n-4\delta_{0,n}\\ nu_n=-\sum_{k=1}^{n}v_{n-k}\,kz_k\\ nv_n=\sum_{k=1}^{n}u_{n-k}\,kz_k\\ (n+1)y_{n+1}=3u_n+\sum_{k=0}^ny_ky_{n-k}-12\delta_{0,n} $$

Lutz Lehmann
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