Solve the initial value problem below using the method of Laplace transforms. y"-4y'+40y=225 e^{5t} y(0)=5 y'(0)=31

he298c 2021-01-31 Answered
Solve the initial value problem below using the method of Laplace transforms.
y"4y+40y=225e5t
y(0)=5
y(0)=31
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Expert Answer

SoosteethicU
Answered 2021-02-01 Author has 102 answers
Step 1
Taking Laplace transform in the both the direction:
y"4y+40y=225e5t
y(0)=5
y(0)=31
L{y"4y+40y}=L{Y(s)}
L{y"}4L{y}+40L{y}=225L{Y(s)}
L{y"}=s2Y(s)sy(0)y(0)
L(y)=sY(s)y(0)
L(y)=Y(s),L{eat}=1(sa)
{s2Y(s)sy(0)y(0)}4{sY(s)y(0)}+40{Y(s)}=225{1(s5)}
{s24s+40}Y(s)+{5s11}=225{1(s5)}
{s24s+40}Y(s)+{5s11}=225{1(s5)}
{s24s+40}Y(s)=225{1(s5)}+5s+11
Y(s)=225{1(s5)}+5s+11s24s+40
Step 2
Find the factor of the function:
Y(s)=225{1(s5)}+5s+11s24s+40=
=225(s5)(s24s+40)+5s(s24s+40)+11(s24s+40)
=225{5s5s24s+40+5(s5)}+5(s2)(s2)2+36+101(s2)2+36+{111(s2)2+36}
Step 3Taking inverse Laplace transform in both direction:
L1{Y(s)}=L1{{(5s5)(s24s+40)+5(s5)}+5(s2)(s2)2+36+101(s2)2+36+{111(s2)2+36}}

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