Elementary Differential Equations

[AST3019]

(dy/dx)^2 + y^2 + 4 = 0; Show that there are no real valued solutions.

(dy/dx)^2 = -(y^2 + 4)
dy/dx= sqrt( -(y^2 + 4)) -----> This is the answer I got, I feel it makes sense because the negative under the radical produce an answer with 'i' or an imaginary sol'n and since y^2 is always positive y^2 + 4> 0, so the there is no way to cancel that negative under the radical.
Unfortunately, this doesn't correspond with the answer in the back of my book, anyone want to let me know what I am doing wrong/ show me a different way to answer it? Thanks.

 

[mighty2000]

Thank you for your post. I would like to offer my explanation for why there are no real valued solutions to this equation.

Firstly, let's rearrange the equation to make it easier to work with:
(dy/dx)^2 = -y^2 - 4

Now, we can see that the left side of the equation is always positive or zero (since it is squared), while the right side is always negative (since y^2 and 4 are always positive). This means that the equation cannot be satisfied for any real value of dy/dx.

Furthermore, if we try to solve for dy/dx by taking the square root of both sides, we will end up with an imaginary solution, as you have correctly pointed out. This is because the square root of a negative number is imaginary.

In summary, there are no real valued solutions to this equation because the left side is always positive or zero, while the right side is always negative. This means that the equation cannot be satisfied for any real value of dy/dx.

I hope this explanation helps clarify things for you. Keep up the good work!