d-2-dt-2-d-dt-0-

Question Number 108094 by Dwaipayan Shikari last updated on 14/Aug/20

\frac{d^{2} Ψ}{{d t}^{2}} + \frac{d Ψ}{d t} + Ψ = 0

Answered by mathmax by abdo last updated on 14/Aug/20

y^{^{″}} + y^{^{'}} + y = 0 \to r^{2} + r + 1 = 0

Δ = 1 - 4 = - 3 \Rightarrow r_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{i \frac{2 π}{3}} and r_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- i \frac{2 π}{3}}

\Rightarrow y = a e^{r_{1} x} + {be}^{r_{2} x} = e^{- \frac{x}{2}} {α \cos (\frac{\sqrt{3}}{2} x) + β \sin (\frac{\sqrt{3}}{2} x)}

Commented by Dwaipayan Shikari last updated on 14/Aug/20

Thanking you ��

Commented by mathmax by abdo last updated on 14/Aug/20

you are welcome