Question

@parthkohli share your method please

Answer 1

\[y''+y'=0\] go

Answer 2

OK, so you kinda see how you can invoke the product rule here.\[1\cdot y'' + 1\cdot y'=0\]Is there a function whose consecutive derivatives are both \(1\)? Obviously not. But what if we find a function whose consecutive derivatives are the same?\[e^x \cdot y'' + e^x \cdot y'=0\]So that's something like the derivative of \(y'e^x\).\[(y'e^x)' = 0\Rightarrow y' e^x = c_1 \Rightarrow y' = c_1 e^{-x} \Rightarrow y = -c_1 e^{-x}+c_2\]

Answer 3

Love it

Answer 4

I am much more simple minded.
\(k^2+k=0\implies k=0\text{ or }k=-1\)
\(y=c_1e^0+c_2e^{-x}\)

Answer 5

Haha yes, I use the same thing @thomas5267 characteristic equation <3