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

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  • $\begingroup$ What is the independent variable? $\endgroup$ May 23 '13 at 21:03
  • $\begingroup$ The independent variable is t $\endgroup$
    – user78885
    May 24 '13 at 0:10
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    $\begingroup$ Is there a good reason to expect there to be a simple general solution? $\endgroup$ May 25 '13 at 20:51
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$\dot{y}+t\cos\dfrac{\pi y}{2}+1-t=y$

$\dfrac{dy}{dt}=\left(1-\cos\dfrac{\pi y}{2}\right)t+y-1$

Let $u=y-1$ ,

Then $y=u+1$

$\dfrac{dy}{dt}=\dfrac{du}{dt}$

$\therefore\dfrac{du}{dt}=\biggl(1-\cos\dfrac{\pi(u+1)}{2}\biggr)t+u$

$\left(\left(1-\cos\left(\dfrac{\pi u}{2}+\dfrac{\pi}{2}\right)\right)t+u\right)\dfrac{dt}{du}=1$

$\left(\left(1+\sin\dfrac{\pi u}{2}\right)t+u\right)\dfrac{dt}{du}=1$

Let $v=t+\dfrac{u}{1+\sin\dfrac{\pi u}{2}}$ ,

Then $t=v-\dfrac{u}{1+\sin\dfrac{\pi u}{2}}$

$\dfrac{dt}{du}=\dfrac{dv}{du}+\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}-\dfrac{1}{1+\sin\dfrac{\pi u}{2}}$

$\therefore\left(1+\sin\dfrac{\pi u}{2}\right)v\left(\dfrac{dv}{du}+\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}-\dfrac{1}{1+\sin\dfrac{\pi u}{2}}\right)=1$

$\left(1+\sin\dfrac{\pi u}{2}\right)v\dfrac{dv}{du}+\left(\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{1+\sin\dfrac{\pi u}{2}}-1\right)v=1$

$\left(1+\sin\dfrac{\pi u}{2}\right)v\dfrac{dv}{du}=\left(1-\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{1+\sin\dfrac{\pi u}{2}}\right)v+1$

$v\dfrac{dv}{du}=\left(\dfrac{1}{1+\sin\dfrac{\pi u}{2}}-\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}\right)v+\dfrac{1}{1+\sin\dfrac{\pi u}{2}}$

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 $v=\dfrac{1}{w}$ ,

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

$\therefore-\dfrac{1}{w^3}\dfrac{dw}{du}=\left(\dfrac{1}{1+\sin\dfrac{\pi u}{2}}-\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}\right)\dfrac{1}{w}+\dfrac{1}{1+\sin\dfrac{\pi u}{2}}$

$\dfrac{dw}{du}=-\dfrac{w^3}{1+\sin\dfrac{\pi u}{2}}+\left(\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}-\dfrac{1}{1+\sin\dfrac{\pi u}{2}}\right)w^2$

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2

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