this is initial value problem.. it is solved by using laplace transform

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y"+y'+9y=0 y'(0)=0, y(0)=0.16

Note that ℒ{y"}=s²Y(s)-sy(0)-y'(0) and ℒ{y'}=sY(s)-y(0).

Also, for s>a, ℒ{eᵃᵗsinwt}=w/((s-a)²+w²) and ℒ{eᵃᵗcost}=(s-a)/((s-a)²+w²).

Apply ℒ to both sides:

s²Y-sy(0)-y'(0)+sY-y(0)+9Y=s²Y-0.16s+sY-0.16+9Y=0 from given initial conditions;

Y(s²+s+9)=0.16(s+1);

Y=0.16(s+1)/(s²+s+9).

Y=0.16[((s+½)+½)/((s+½)²+8.75)];

Y=0.16[(s+½)/((s+½)²+8.75)+(√8.75/17.5)(√8.75/((s+½)²+8.75)].

(These steps arrange the expression into a form which lends itself to converting back from Laplace Transform.)

Apply inverse ℒ to both sides:

y(t)=0.16[(e^-(t/2))cos(t√8.75)+(√8.75/17.5)(e-(t/2))sin(t√8.75)].

by Top Rated User (695k points)

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