Consider the differential equation-d2ytdt2-2dytdt-yt- t with y t -t-0-2 and dydt-t-0-0- The numerical value of dydt-t-0- is

D^2y ( t )dt^2+2dy ( t )dt+y(t)=δ ( t ) ⇒ s^2Y ( s ) sy(0) y'(0)+2[sY(s) y(0)]+Y(s)=1 ⇒ Y(s) [ s^2+2s+1 ] ( s× 2 ) y'(0) (2× 2) ⇒ Y(s) [ s^2+2s+1 ]= ...

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Consider the ...

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$Consider the differential equation:$ $\frac{d^{2} y (t)}{d t^{2}} + 2 \frac{d y (t)}{d t} + y (t) = δ (t)$ $with$ $y (t) |_{t = 0^{-}} = - 2$ and $\frac{d y}{d t} |_{t = 0^{-}} = 0.$ $The numerical value of \frac{d y}{d t} |_{t = 0^{+}} is$

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$\frac{d^{2} y (t)}{d t^{2}} + 2 \frac{d y (t)}{d t} + y (t) = δ (t)$

$\Rightarrow s^{2} Y (s) - s y (0) - y^{'} (0) + 2 [s Y (s) - y (0)] + Y (s) = 1$

$\Rightarrow Y (s) [s^{2} + 2 s + 1] - (s \times - 2) - y^{'} (0) - (2 \times 2)$

$\Rightarrow Y (s) [s^{2} + 2 s + 1] = - 2 s - 3$

$\Rightarrow Y (s) = \frac{- 2 s - 3}{s^{2} + 2 s + 1} = \frac{- 2}{s + 1} - \frac{1}{{(s + 1)}^{2}}$

$\Rightarrow y (t) = - 2 e^{- t} - t e^{- t}$

$\frac{d y (t)}{d t} = 2 e^{- t} - (- t e^{- t} + e^{- t})$

$= e^{- t} + t e^{- t}$

$\frac{d y (t)}{d t} |_{t = 0^{+}} = 1 + 0 = 1$

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Consider the differential equation:d2y(t)dt2+2dy(t)dt+y(t)=δ(t) with y(t)|t=0−=−2 and dydt|t=0−=0. The numerical value of dydt|t=0+ is

d2y(t)dt2+2dy(t)dt+y(t)=δ(t) ⇒s2Y(s)−sy(0)−y′(0)+2[sY(s)−y(0)]+Y(s)=1 ⇒Y(s)[s2+2s+1]−(s×−2)−y′(0)−(2×2) ⇒Y(s)[s2+2s+1]=−2s−3 ⇒Y(s)=−2s−3s2+2s+1=−2s+1−1(s+1)2 ⇒y(t)=−2e−t−te−t dy(t)dt=2e−t−(−te−t+e−t) =e−t+te−t dy(t)dt|t=0+=1+0=1

$Consider the differential equation:$ $\frac{d^{2} y (t)}{d t^{2}} + 2 \frac{d y (t)}{d t} + y (t) = δ (t)$ $with$ $y (t) |_{t = 0^{-}} = - 2$ and $\frac{d y}{d t} |_{t = 0^{-}} = 0.$ $The numerical value of \frac{d y}{d t} |_{t = 0^{+}} is$