Non-linear first order differential equation 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.
 A: $\dot{y}+t\cos\dfrac{\pi y}{2}+1-t=y$
$\dfrac{dy}{dt}=\left(1-\cos\dfrac{\pi y}{2}\right)t+y-1$
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
