Differential equation

[tiny leaf]

I know that this DE can be solved in 3 different ways

For ∆ = b^2 − 4c > 0 -> Kexp(-rt)
For ∆ = b^2 − 4c = 0 -> ...
For ∆ = b^2 − 4c < 0 -> Kexp(−b±i√−∆^2t)

[strange elm]

A lot of the notation is really confusing. For one thing, (E) is not defined. I'll assume it's a marker referring to the DE. There are a few other problems, but I can at least tell you that what it's asking you to do is: Prove that Cexp(rt) is a solution of the DE. I don't know what I is here, and the notation that $t\in I\to C\exp{rt}$ is hard to parse, so I can't speak on any of the details. But if you just kind of plug your ears and compute anyway, you get

[waxen musk]

addem

[strange elm]

$$ (C\exp{rt})'' + b(C\exp{rt})' + cC\exp{rt} = r^2 C\exp{rt}+brC\exp{rt} + cC\exp{rt}$$

I'm not sure why that cut off in the TeXit.

I guess the line is too long.

$$ (C\exp{rt})'' + b(C\exp{rt})' + cC\exp{rt}$$

[waxen musk]

addem

[strange elm]

Equals $$r^2 C\exp{rt}+brC\exp{rt} + cC\exp{rt} $$

[waxen musk]

addem

[strange elm]

Which is the same as $$ C\exp{rt}(r^2+br+c) $$. If you now use what you know about $r$ you should get that this is equal to zero, and that will prove that indeed $C\exp{rt}$ is a solution for the DE.