ntaraxq

2022-07-03

I've found this particular equation rather tough, can you give me some hints on how to solve
$\stackrel{˙}{y}+t\mathrm{cos}\frac{\pi y}{2}+1-t=y$

Zachery Conway

Expert

$\stackrel{˙}{y}+t\mathrm{cos}\frac{\pi y}{2}+1-t=y$
$\frac{dy}{dt}=\left(1-\mathrm{cos}\frac{\pi y}{2}\right)t+y-1$$\stackrel{˙}{y}+t\mathrm{cos}\frac{\pi y}{2}+1-t=y$
Let $u=y-1$,
Then $y=u+1$
$\frac{dy}{dt}=\frac{du}{dt}$
$\therefore \frac{du}{dt}=\left(1-\mathrm{cos}\frac{\pi \left(u+1\right)}{2}\right)t+u$
$\left(\left(1-\mathrm{cos}\left(\frac{\pi u}{2}+\frac{\pi }{2}\right)\right)t+u\right)\frac{dt}{du}=1$
$\left(\left(1+\mathrm{sin}\frac{\pi u}{2}\right)t+u\right)\frac{dt}{du}=1$
Let $v=t+\frac{u}{1+\mathrm{sin}\frac{\pi u}{2}}$,
Then $t=v-\frac{u}{1+\mathrm{sin}\frac{\pi u}{2}}$
$\frac{dt}{du}=\frac{dv}{du}+\frac{\frac{\pi u}{2}\mathrm{cos}\frac{\pi u}{2}}{{\left(1+\mathrm{sin}\frac{\pi u}{2}\right)}^{2}}-\frac{1}{1+\mathrm{sin}\frac{\pi u}{2}}$
$\therefore \left(1+\mathrm{sin}\frac{\pi u}{2}\right)v\left(\frac{dv}{du}+\frac{\frac{\pi u}{2}\mathrm{cos}\frac{\pi u}{2}}{{\left(1+\mathrm{sin}\frac{\pi u}{2}\right)}^{2}}-\frac{1}{1+\mathrm{sin}\frac{\pi u}{2}}\right)=1$
$\left(1+\mathrm{sin}\frac{\pi u}{2}\right)v\frac{dv}{du}+\left(\frac{\frac{\pi u}{2}\mathrm{cos}\frac{\pi u}{2}}{1+\mathrm{sin}\frac{\pi u}{2}}-1\right)v=1$
$\left(1+\mathrm{sin}\frac{\pi u}{2}\right)v\frac{dv}{du}=\left(1-\frac{\frac{\pi u}{2}\mathrm{cos}\frac{\pi u}{2}}{1+\mathrm{sin}\frac{\pi u}{2}}\right)v+1$
$v\frac{dv}{du}=\left(\frac{1}{1+\mathrm{sin}\frac{\pi u}{2}}-\frac{\frac{\pi u}{2}\mathrm{cos}\frac{\pi u}{2}}{{\left(1+\mathrm{sin}\frac{\pi u}{2}\right)}^{2}}\right)v+\frac{1}{1+\mathrm{sin}\frac{\pi u}{2}}$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $v=\frac{1}{w}$,
Then $\frac{dv}{du}=-\frac{1}{{w}^{2}}\frac{dw}{du}$
$\therefore -\frac{1}{{w}^{3}}\frac{dw}{du}=\left(\frac{1}{1+\mathrm{sin}\frac{\pi u}{2}}-\frac{\frac{\pi u}{2}\mathrm{cos}\frac{\pi u}{2}}{{\left(1+\mathrm{sin}\frac{\pi u}{2}\right)}^{2}}\right)\frac{1}{w}+\frac{1}{1+\mathrm{sin}\frac{\pi u}{2}}$
$\frac{dw}{du}=-\frac{{w}^{3}}{1+\mathrm{sin}\frac{\pi u}{2}}+\left(\frac{\frac{\pi u}{2}\mathrm{cos}\frac{\pi u}{2}}{{\left(1+\mathrm{sin}\frac{\pi u}{2}\right)}^{2}}-\frac{1}{1+\mathrm{sin}\frac{\pi u}{2}}\right){w}^{2}$

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