ntaraxq

Answered

2022-07-03

I've found this particular equation rather tough, can you give me some hints on how to solve

$\dot{y}+t\mathrm{cos}\frac{\pi y}{2}+1-t=y$

$\dot{y}+t\mathrm{cos}\frac{\pi y}{2}+1-t=y$

Answer & Explanation

Zachery Conway

Expert

2022-07-04Added 7 answers

$\dot{y}+t\mathrm{cos}{\displaystyle \frac{\pi y}{2}}+1-t=y$

$\frac{dy}{dt}}=(1-\mathrm{cos}{\displaystyle \frac{\pi y}{2}})t+y-1$$\dot{y}+t\mathrm{cos}{\displaystyle \frac{\pi y}{2}}+1-t=y$

Let $u=y-1$,

Then $y=u+1$

$\frac{dy}{dt}}={\displaystyle \frac{du}{dt}$

$\therefore {\displaystyle \frac{du}{dt}}={\textstyle (}1-\mathrm{cos}{\displaystyle \frac{\pi (u+1)}{2}}{\textstyle )}t+u$

$((1-\mathrm{cos}({\displaystyle \frac{\pi u}{2}}+{\displaystyle \frac{\pi}{2}}))t+u){\displaystyle \frac{dt}{du}}=1$

$((1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})t+u){\displaystyle \frac{dt}{du}}=1$

Let $v=t+{\displaystyle \frac{u}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}$,

Then $t=v-{\displaystyle \frac{u}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}$

$\frac{dt}{du}}={\displaystyle \frac{dv}{du}}+{\displaystyle \frac{{\displaystyle \frac{\pi u}{2}}\mathrm{cos}{\displaystyle \frac{\pi u}{2}}}{{(1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})}^{2}}}-{\displaystyle \frac{1}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}$

$\therefore (1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})v({\displaystyle \frac{dv}{du}}+{\displaystyle \frac{{\displaystyle \frac{\pi u}{2}}\mathrm{cos}{\displaystyle \frac{\pi u}{2}}}{{(1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})}^{2}}}-{\displaystyle \frac{1}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}})=1$

$(1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})v{\displaystyle \frac{dv}{du}}+({\displaystyle \frac{{\displaystyle \frac{\pi u}{2}}\mathrm{cos}{\displaystyle \frac{\pi u}{2}}}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}-1)v=1$

$(1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})v{\displaystyle \frac{dv}{du}}=(1-{\displaystyle \frac{{\displaystyle \frac{\pi u}{2}}\mathrm{cos}{\displaystyle \frac{\pi u}{2}}}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}})v+1$

$v{\displaystyle \frac{dv}{du}}=({\displaystyle \frac{1}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}-{\displaystyle \frac{{\displaystyle \frac{\pi u}{2}}\mathrm{cos}{\displaystyle \frac{\pi u}{2}}}{{(1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})}^{2}}})v+{\displaystyle \frac{1}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}$

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 $v={\displaystyle \frac{1}{w}}$,

Then $\frac{dv}{du}}=-{\displaystyle \frac{1}{{w}^{2}}}{\displaystyle \frac{dw}{du}$

$\therefore -{\displaystyle \frac{1}{{w}^{3}}}{\displaystyle \frac{dw}{du}}=({\displaystyle \frac{1}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}-{\displaystyle \frac{{\displaystyle \frac{\pi u}{2}}\mathrm{cos}{\displaystyle \frac{\pi u}{2}}}{{(1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})}^{2}}}){\displaystyle \frac{1}{w}}+{\displaystyle \frac{1}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}$

$\frac{dw}{du}}=-{\displaystyle \frac{{w}^{3}}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}+({\displaystyle \frac{{\displaystyle \frac{\pi u}{2}}\mathrm{cos}{\displaystyle \frac{\pi u}{2}}}{{(1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})}^{2}}}-{\displaystyle \frac{1}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}){w}^{2$

$\frac{dy}{dt}}=(1-\mathrm{cos}{\displaystyle \frac{\pi y}{2}})t+y-1$$\dot{y}+t\mathrm{cos}{\displaystyle \frac{\pi y}{2}}+1-t=y$

Let $u=y-1$,

Then $y=u+1$

$\frac{dy}{dt}}={\displaystyle \frac{du}{dt}$

$\therefore {\displaystyle \frac{du}{dt}}={\textstyle (}1-\mathrm{cos}{\displaystyle \frac{\pi (u+1)}{2}}{\textstyle )}t+u$

$((1-\mathrm{cos}({\displaystyle \frac{\pi u}{2}}+{\displaystyle \frac{\pi}{2}}))t+u){\displaystyle \frac{dt}{du}}=1$

$((1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})t+u){\displaystyle \frac{dt}{du}}=1$

Let $v=t+{\displaystyle \frac{u}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}$,

Then $t=v-{\displaystyle \frac{u}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}$

$\frac{dt}{du}}={\displaystyle \frac{dv}{du}}+{\displaystyle \frac{{\displaystyle \frac{\pi u}{2}}\mathrm{cos}{\displaystyle \frac{\pi u}{2}}}{{(1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})}^{2}}}-{\displaystyle \frac{1}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}$

$\therefore (1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})v({\displaystyle \frac{dv}{du}}+{\displaystyle \frac{{\displaystyle \frac{\pi u}{2}}\mathrm{cos}{\displaystyle \frac{\pi u}{2}}}{{(1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})}^{2}}}-{\displaystyle \frac{1}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}})=1$

$(1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})v{\displaystyle \frac{dv}{du}}+({\displaystyle \frac{{\displaystyle \frac{\pi u}{2}}\mathrm{cos}{\displaystyle \frac{\pi u}{2}}}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}-1)v=1$

$(1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})v{\displaystyle \frac{dv}{du}}=(1-{\displaystyle \frac{{\displaystyle \frac{\pi u}{2}}\mathrm{cos}{\displaystyle \frac{\pi u}{2}}}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}})v+1$

$v{\displaystyle \frac{dv}{du}}=({\displaystyle \frac{1}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}-{\displaystyle \frac{{\displaystyle \frac{\pi u}{2}}\mathrm{cos}{\displaystyle \frac{\pi u}{2}}}{{(1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})}^{2}}})v+{\displaystyle \frac{1}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}$

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 $v={\displaystyle \frac{1}{w}}$,

Then $\frac{dv}{du}}=-{\displaystyle \frac{1}{{w}^{2}}}{\displaystyle \frac{dw}{du}$

$\therefore -{\displaystyle \frac{1}{{w}^{3}}}{\displaystyle \frac{dw}{du}}=({\displaystyle \frac{1}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}-{\displaystyle \frac{{\displaystyle \frac{\pi u}{2}}\mathrm{cos}{\displaystyle \frac{\pi u}{2}}}{{(1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})}^{2}}}){\displaystyle \frac{1}{w}}+{\displaystyle \frac{1}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}$

$\frac{dw}{du}}=-{\displaystyle \frac{{w}^{3}}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}+({\displaystyle \frac{{\displaystyle \frac{\pi u}{2}}\mathrm{cos}{\displaystyle \frac{\pi u}{2}}}{{(1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}})}^{2}}}-{\displaystyle \frac{1}{1+\mathrm{sin}{\displaystyle \frac{\pi u}{2}}}}){w}^{2$

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