how do i do this. calc (2 Viewers)

yashbb · Sep 29, 2021

sorry for bad image quality

Drongoski · Sep 29, 2021

yashbb · Sep 29, 2021

Drongoski said:
$y = e^{kx} \implies y' = ke^{kx} \implies y'' = k^2e^{kx}\\ \\ \therefore k^2e^{kx} + 7ke^{kx} + 12e^{kx} = 0$

i understand that part but i dont know what to do after, sorry should have specified.

CM_Tutor · Sep 29, 2021

$y=e^{kx} \qquad \implies \qquad \cfrac{dy}{dx} = ke^{kx} \qquad \implies \qquad \cfrac{d^2y}{dx^2} = k^2e^{kx}$

Substitute these into the given equation:

$\begin{align*} \cfrac{d^2y}{dx^2} + 7\cfrac{dy}{dx} + 12y &= 0 \\ k^2e^{kx} + 7ke^{kx} + 12e^{kx} &= 0 \\ \left(k^2 + 7k + 12\right)e^{kx} &= 0 \\ \(k + 3)(k + 4)e^{kx} &= 0 \end{align*}$

There are two possible solutions, $k = -3$ and $k = -4$ , as $e^{kx} > 0$ for all $x$ , no matter the value of $k$ .

yashbb · Sep 29, 2021

CM_Tutor said:
$y=e^{kx} \qquad \implies \qquad \cfrac{dy}{dx} = ke^{kx} \qquad \implies \qquad \cfrac{d^2y}{dx^2} = k^2e^{kx}$

Substitute these into the given equation:

$\begin{align*} \cfrac{d^2y}{dx^2} + 7\cfrac{dy}{dx} + 12y &= 0 \\ k^2e^{kx} + 7ke^{kx} + 12e^{kx} &= 0 \\ \left(k^2 + 7k + 12\right)e^{kx} &= 0 \\ \(k + 3)(k + 4)e^{kx} &= 0 \end{align*}$

There are two possible solutions, $k = -3$ and $k = -4$ , as $e^{kx} > 0$ for all $x$ , no matter the value of $k$ .

ahh i see, thank you!

how do i do this. calc (2 Viewers)

yashbb

Active Member

Drongoski

Well-Known Member

yashbb

Active Member

CM_Tutor

Moderator

yashbb

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)