general solution of ODE?

Question

There is this differential equation with I can't solve. I'm worried it will show up again on the exam. Supposedly, it is solved my substituting v=sin(y) but I don't see how cos(y) goes away. Here it is:

(2xsin(y)cos(y))y' = 4x^2 + sin^2(y)

After subbing v=sin(y), the book says you have:
2xvv' = 4x^2 + v^2

My only question is, how the heck did cosine go away, and why is there magically a v' ???

Thanks! this has had me stumped all day.

Answer 1

Your questions are linked. If v = sin(y), then what is v' ?

Chain Rule gives:
v' = cos(y)y'

(Did you notice the y' gone as well?)

Voila! (Good Luck!)

Answer 2

Well... what is the derivative of v = sin(y) with respect to x? Since the derivative of sine is cosine, taking the derivative of sin(y) gave you v' = cos(y)*y'. So, they just noticed that cos(y)*y' is equal to v', so they plugged that in there.

So basically, the right side of your equation is essentially the same, except that they plugged in v for sin(y), and they did the same thing on the left side, while also plugging in v' where there was a cos(y)*y'.

Answer 3

if v = sin(y), then v' is cos(y)*y'. You take that term combination cos(y)*y' and substitute v' for it.