Question

Show that the equation (dy/dx)^2 + y^2 +4 = 0 has a no (real-valued) solution.

Answer 1

I'm guessing you can just treat y' (u don't mean y''?) as any other variable here so

y' = sqrt (-(y^2+4)) ie no real solutions.

Answer 2

because prefect squares are non-negative so you cannot add 4 to them and get 0

Answer 3

i know nothing of differential equations, but seems clear anyway yes?

Answer 4

Ya, it makes sense ha. What I wrote is how I meant it. I was just confused about how the back of the book solved the problem. I like your way better ha. This was their solution: (dy/dx)^2 + y^2 >= 0 and we have (dy/dx)^2 + y^2 +4 >= 4. Hence, it can never be zero....it's the same idea I guess, just an indirect way of stating it.