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svp [43]
3 years ago
9

Use Laplace transforms to solve the following initial value problem: x"+8x'+15x = 0; x(0) = 2, x'(0) = -3 PLEASE SHOW ALL WORK,

OR RISK LOSING ALL POINTS!!!! x')=sX (s) - x(0) x"(t) = sº X(s) - sx(0) - x'(0)
Mathematics
1 answer:
Lisa [10]3 years ago
8 0

Taking the transform of both sides gives

\mathcal L_s\{x''+8x'+15x\}=0

(s^2X(s)-sx(0)-x'(0))+8(sX(s)-x(0))+15X(s)=0

where X(s) denotes the Laplace transform of x(t), \mathcal L_s\{x(t)\}. Solve for X(s) to get

(s^2+8s+15)X(s)=2s+13

X(s)=\dfrac{2s+13}{s^2+8s+15}=\dfrac{2s+13}{(s+3)(s+5)}

Split the right side into partial fractions:

\dfrac{2s+13}{(s+3)(s+5)}=\dfrac a{s+3}+\dfrac b{s+5}

2s+13=a(s+5)+b(s+3)

If s=-3, then 7=2a\implies a=\dfrac72; if s=-5, then 3=-2b\implies b=-\dfrac32. So

X(s)=\dfrac72\dfrac1{s+3}-\dfrac32\dfrac1{s+5}

Finally, take the inverse transform of both sides to solve for x(t):

x(t)=\dfrac72e^{-5t}-\dfrac32e^{-3t}

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