19″She walks through the hallways, holding a box of props, towards the auditorium, when she spots you alone, surprised. She waltzes up to you with a welcoming smile. Oh, I didn't think they were serious with that whole desegregation thingy! Welcome to the school. I didn't think I'd see another predator in person besides my two wolf dads... She shrugs casually with a smile. Long story. Anyways, I'm Cameronnie. Need someone to show you around? I won't mind!
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3Talkior-meXBiSoy
25/04/2025
*he grabbed a towel and the icepack and pressed it against it*
To solve the differential equation \(3y'' + 3y' - 6y = 0\) with initial conditions \(y(0) = 7\) and \(y'(0) = 9\): 1. **Find the characteristic equation**: \[ 3r^2 + 3r - 6 = 0 \quad \Rightarrow \quad r^2 + r - 2 = 0 \quad (\text{after dividing by } 3) \] 2. **Solve the characteristic equation**: \[ r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2(1)} = \frac{-1 \pm 3}{2} \quad \Rightarrow \quad r = 1 \text{ and } r = -2 \] 3. **General solution**: \[ y(t) = C_1 e^{t} + C_2 e^{-2t} \] 4. **Apply initial conditions**: - \(y(0) = 7\): \[ C_1 + C_2 = 7 \] - \(y'(0) = 9\): \[ C_1 - 2C_2 = 9 \] 5. **Solve for constants \(C_1\) and \(C_2\)**: - Subtract the first equation from the second: \[ (C_1 - 2C_2) - (C_1 + C_2) = 9 - 7 \quad \Rightarrow \quad -3C_2 = 2 \quad \Rightarrow \quad C_2 = -\frac{2}{3} \] - Substitute \(C_2 = -\frac{2}{3}\) into \(C_1 + C_2 = 7\): \[ C_1 - \frac{2}{3} = 7 \quad \Rightarrow \quad C_1 = \frac{23}{3} \] 6. **Final solution**: \[ y(t) = \frac{23}{3} e^{t} - \frac{2}{3} e^{-2t} \] \boxed{\dfrac{23}{3} e^{t} - \dfrac{2}{3} e^{-2 t}}
From the memory
2 Memories
Talkior-meXBiSoy
25/04/2025