Solve the third-order initial value problem below using the method of Laplace transforms y'''+3y''−18y'−40y=−120 y(0)=6 y'(0)=45 y"(0)=-21

Efan Halliday 2021-03-02 Answered
Solve the third-order initial value problem below using the method of Laplace transforms
y+3y18y40y=120
y(0)=6
y(0)=45
y"(0)=21
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Expert Answer

Delorenzoz
Answered 2021-03-03 Author has 91 answers
Step 1
Given equation,
y+3y18y40y=120
y(0)=6
y(0)=45
y"(0)=21
take Laplace transform of both sides of the equation (1)
Step 2L{y+3y18y40y}=L{120}
s3L{y}s2y(0)sy(0)y(0)+3(s2L{y}sy(0)y(0))
18(sL{y}y(0))40L{y}=120s
Plug in the initial conditions: y(0)=6,y(0)=45,y"(0)=21
s3L{y}s2×6s×45(21)+3(s2L{y}s×645)18(sL{y}6)40L{y}=120s
s3L{y}6s245s+21+3(s2L{y}6s45)18(sL{y}6)40L{y}=120s
s3L{y}+3s2L{y}18sL{y}40L{y}6s263s6=120s
L{y}(s3+3s218s40)=120s+6s2+63s+6
L{y}(s3+3s218s40)=120+6s3+63s2+6ss
L{y}=120+6s3+63s2+6ss(s3+3s218s40)
Step 3
Resolve into partial fractions,
6s3+63s2+6s120s(s3+3s218s40)=6s3+63s2+6s120s(s+2)(s4)(s+5)
=AsBs+2Cs4Ds+5

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