3(d^2 y/dx^2 )+4(dy/dx)+5y=0 y=?


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Question Number 116055 by Study last updated on 30/Sep/20

$3 \frac{d^{2} y}{{d x}^{2}} + 4 \frac{d y}{d x} + 5 y = 0 y = ?$
Commented by mohammad17 last updated on 30/Sep/20
$(3D^2 +4D+5)y=0 (3m^2 +4m+5)=0 m=((−4±(√(16−60)))/6)⇒m_1 =−(2/3)−((√(44))/6)i and m_2 =−(2/3)+((√(44))/6)i y=e^(−(2/3)x) {C_1 cos(((√(44))/6)x)+C_2 sin(((√(44))/6)x)} ≪m.o≫$
$(3 D^{2} + 4 D + 5) y = 0$ $(3 m^{2} + 4 m + 5) = 0$ $m = \frac{- 4 \pm \sqrt{16 - 60}}{6} \Rightarrow m_{1} = - \frac{2}{3} - \frac{\sqrt{44}}{6} i a n d m_{2} = - \frac{2}{3} + \frac{\sqrt{44}}{6} i$ $y = e^{- \frac{2}{3} x} {C_{1} c o s (\frac{\sqrt{44}}{6} x) + C_{2} s i n (\frac{\sqrt{44}}{6} x)}$ $≪ m . o ≫$
	Answered by Bird last updated on 30/Sep/20
	$3y^(′′) +4y^′ +5=0 ⇒3r^2 +4r +5=0 Δ^(′ ) =4−15 =−11 ⇒ r_1 =((−2+i(√(11)))/3) and r_2 =((−2−i(√(11)))/3) ⇒y =α e^(r_1 x) +βe^(r_2 x) =e^((−2x)/3) {acos(((√(11))/3)x)+bsin(((√(11))/3)x)}$
	$3 y^{^{″}} + 4 y^{^{'}} + 5 = 0 \Rightarrow 3 r^{2} + 4 r + 5 = 0$ $Δ^{^{'}} = 4 - 15 = - 11 \Rightarrow$ $r_{1} = \frac{- 2 + i \sqrt{11}}{3} a n d r_{2} = \frac{- 2 - i \sqrt{11}}{3}$ $\Rightarrow y = α e^{r_{1} x} + β e^{r_{2} x} = e^{\frac{- 2 x}{3}} {a c o s (\frac{\sqrt{11}}{3} x) + b s i n (\frac{\sqrt{11}}{3} x)}$
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