Question #110910
(d^2 y)/(dt^2) - 2 dy/dt + 1=0 with y(t=0)=5 and y ̇(t=0)=-9
use laplace transform
1
Expert's answer
2020-04-24T20:16:23-0400

Let's take the transform of every term in the differential equation

L[y]L[2y]+L[1]=0L[y'']-L[2y']+L[1]=0

It is known, that

L[y]=s2L[y]sy(0)y(0)=s2Y(s)5s+9L[y'']=s^2L[y]-sy(0)-y'(0)=s^2Y(s)-5s+9

L[y]=sY(s)y(0)=sY(s)5L[y']=sY(s)-y(0)=sY(s)-5 .

Here Y(s)=L[y]Y(s)=L[y]

The Laplace-transformed differential equation is

s2Y(s)5s+92(sY(s)5)+1s=0s^2Y(s)-5s+9-2(sY(s)-5)+\frac{1}{s}=0

Y(s22s)=5s1s19Y(s^2-2s)=5s-\frac{1}{s}-19

Y=5s219s1s(s22s)=12s2+394s194(s2)Y=\frac{5s^2-19s-1}{s(s^2-2s)}=\frac{1}{2s^2}+\frac{39}{4s}-\frac{19}{4(s-2)}

Recalling the inverse transform will lead us to

1s2x;1s1;1s2e2t\frac{1}{s^2}\gets x;\quad \frac{1}{s}\gets 1;\quad \frac{1}{s-2}\gets e^{2t}

Therefore

Yy=t2+394194e2tY\gets y=\frac{t}{2}+\frac{39}{4}-\frac{19}{4}e^{2t}


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