Is the function $y(t)$ is a solution of the equation $y'=\sin(yt)$? Is the function $y(t)$ a solution of the equation $y'=\sin(yt)$?
any thought to start me up?
I'm not sure what is the question asking.
EDIT: 
Someone tell me if I'm correct or not . 
If I'm finding the general solution of the equation y'=ty, does this mean I'm finding the anti-derivative of that which is y=((t(y^2))/2)+C ? 
Very confused at what the question want. 
 A: I think the question is: is there a function $y(t)$ such that $$\frac{\mathbb d}{\mathbb d t}y(t) = \sin{ty(t)}$$ $y(t) = 0$ certainly works. Less trivial functions, I'm pretty sure do not.
A: One way to start this is to think of what $y$ is related to. That is, it's a function of time. Meanwhile, $sin(yt)$ is also a function of time and also a function of y. 
To get the solution to $y' = \sin(yt)$ one can ask what is the derivative of sine and cosine? In this case you can't separate variables. But you can assume that a general solution will look roughly like this: $A\sin(yt) + B\cos(yt)$. (I guess you could have exopnential functions but for this kind of problem that's unnecessarily harder to do). 
Also the antiderivative you have isn't quite right. If you have $y' = ty$ then it's a separable equation. That means you can just move stuff to the left. So you would rearrange that as $\frac{y'}{y} = t$. 
But that's an unwieldy notation to use. I would have set it up as follows: 
$$y' = \frac{dy}{dt} = ty \rightarrow \frac{dy}{y} = tdt$$ 
That one you can integrate both sides and end up with 
$\ln y = \frac{t^2}{2} + C$
And you have an exponential function to get y. $e^{\ln y} = e^{\frac{t^2}{2}+ C} \rightarrow y = e^C e^{\frac{t^2}{2}} \rightarrow A e^{\frac{t^2}{2}}$
(Because e raised to any constant power is a constant too)
And you can check this by taking a derivative and plugging it into the $y'=ty$. 
I hope this helps. 
A: If you set $v = ty$ then $y = \frac{v}{t}$ so the equation becomes $$(\frac{v}{t})' = \frac{v'}{t} - \frac{v}{t^2} = \sin(v)$$ WolframAlpha didn't give a solution though...
A: Let $u=ty$ ,
Then $y=\dfrac{u}{t}$ 
$\dfrac{dy}{dt}=\dfrac{1}{t}\dfrac{du}{dt}-\dfrac{u}{t^2}$
$\therefore\dfrac{1}{t}\dfrac{du}{dt}-\dfrac{u}{t^2}=\sin u$
$\dfrac{1}{t}\dfrac{du}{dt}=\dfrac{t^2\sin u+u}{t^2}$
$(t^2\sin u+u)\dfrac{dt}{du}=t$
Let $v=t^2$ ,
$\dfrac{dv}{du}=2t\dfrac{dt}{du}$
$\therefore\dfrac{(t^2\sin u+u)}{2t}\dfrac{dv}{du}=t$
$(t^2\sin u+u)\dfrac{dv}{du}=2t^2$
$(v\sin u+u)\dfrac{dv}{du}=2v$
Let $w=v+u\csc u$ ,
Then $v=w-u\csc u$
$\dfrac{dv}{du}=\dfrac{dw}{du}+(u\cot u-1)\csc u$
$\therefore(\sin u)w\left(\dfrac{dw}{du}+(u\cot u-1)\csc u\right)=2(w-u\csc u)$
$(\sin u)w\dfrac{dw}{du}+(u\cot u-1)w=2w-2u\csc u$
$(\sin u)w\dfrac{dw}{du}=(3-u\cot u)w-2u\csc u$
$w\dfrac{dw}{du}=(3\csc u-u\csc u\cot u)w-2u\csc^2u$
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 $w=\dfrac{1}{z}$ ,
Then $\dfrac{dw}{du}=-\dfrac{1}{z^2}\dfrac{dz}{du}$
$\therefore-\dfrac{1}{z^3}\dfrac{dz}{du}=\dfrac{3\csc u-u\csc u\cot u}{z}-2u\csc^2u$
$\dfrac{dz}{du}=2z^3u\csc^2u+(u\csc u\cot u-3)z^2$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
