Use laplace transform to solve the given system a) frac{dx}{dt} -2x- frac{dx}{dt}-y =6e^{3t} 2left(frac{dx}{dt}right)-3x+frac{dx}{dt}-3y=6e^{3t} x(0)=3 y(0)=0

amanf 2021-01-06 Answered
Use laplace transform to solve the given system
a)dxdt2xdxdty=6e3t
2(dxdt)3x+dxdt3y=6e3tx(0)=3y(0)=0
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Expert Answer

odgovoreh
Answered 2021-01-07 Author has 107 answers
Step 1
Given
a)dxdt2xdxdty=6e3t
2(dxdt)3x+dxdt3y=6e3tx(0)=3y(0)=0
We use the Laplace transforms initially,
L(x(t))=x¯(s)
L(y(t))=y¯(s)
L(x(t))=sx¯(s)x(0)
L(y(t))=sy¯(s)y(0)
And in the latter stage, we use the known inverse Laplace transforms to get the required solution.
Step 2
So, we have
sx¯(s)x(0)2x¯(s)sy¯(s)+y(0)+y¯(s)=6(s3)
2sx¯(s)2x(0)3x¯(s)+sx¯(s)y(0)3y¯(s)=6(s3)
Simplifying and applying the initial conditions,
(s2)x¯(s)+(1s)y¯(s)=6(s3)+3(1)
(2s3)x¯(s)+(x3)y¯(s)=6(s3)+3(2)
Using Cramer's rule,
x¯(s)=|6(s3)+31s6(s3)+6s3||s21s2s3s3|
=3s36(s2)(1s)s3(s2)(s3)(2s3)(1s)
=(s1)(9s21)3s210s+9(3)
=363s210s+9
=32(s106)2+(86)2
x(t)=3δ(t)6/8e10t6sin(8t6)
Where, δ(t) is the delta function
Step 3
Substituting (3) in (2), we get
(2s3)(s1)(9s21)3s210s+9+(s3)y¯(s)=6(s3)+6

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