Question

Show that equation (coming soon) has no (real-valued) solution.

Answer 1

\[(dy/dx)^{2}+y ^{2}+4=0\]

Answer 2

Can you re write it like\[\frac{ d ^{2}y }{ dx ^{2}}\]

Answer 3

i think

\[\left(\frac{ \text d y }{ \text dx }\right)^2=y'^2\]
is not the same as
\[\frac{ \text d^2 y }{ \text dx^2 }=y''\]

Answer 4

Does anybody know what my first step might be?

Answer 5

find dy/dx

Answer 6

so solve for it.

Answer 7

\[\frac{ dy }{ dx }=\sqrt{-y ^{2}-4}\]

Answer 8

So already I can see that im going to have i in the answer. But i still need to prove it

Answer 9

\[\sqrt{-1(x ^{2}+4)}\]

Answer 10

\[i \sqrt{x ^{2}+4}\]Do you think that should be enough to satisfy the question

Answer 11

@UnkleRhaukus plx take a look at my soln

Answer 12

i just saw that i wrote c for y but if you ignore that am i correct

Answer 13

x***

Answer 14

what you are doing makes sense to me , can you factor x^2 +4 ?

Answer 15

No

Answer 16

hmm

Answer 17

I guess this is it. Thanks for helping me get started. I probably just had to show that i is in the solution.

Answer 18

(x+i2)(x-i2)=x^2+4

Answer 19

I didnt think of that. nice

Answer 20

i dont know what kind of answer they are looking for , i havent this type of question before

Answer 21

I think we satisfied it by proving that the solution is riddled with the nonreal number i