Solve the following differential equation by the laplace transform method. 6y^{(4)}+7y'''+21y"+28y'-12y=t(\cos(2t)+te^{\frac{-3t}{2}}).

Chardonnay Felix 2021-09-03 Answered

Solve the following differential equation by the laplace transform method
6y(4)+7y+21y28y12y=t(cos(2t)+te3t2)
y(0)=0,y(0)=1,y"(0)=0,y(0)=1

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Expert Answer

Ezra Herbert
Answered 2021-09-04 Author has 99 answers

Step 1
Given equation is
6y(4)+7y+21y28y12y=t(cos(2t)+te3t2)
Apply Laplace transform on both sides
L{6y(4)+7y+21y"+28y12y}=L{t(cos(2t)+te3t2)}
6L{y(t)}+7L{y(t)}+21L{y"(t)}+28L{y(t)}+12L{y}=L{t(cos(2t)+te3t2)}
6(s4L{y}s3y(0)s2y(0)sy"(0)y(0))+7(s3L{y}s2y(0)sy(0)y"(0))+21(s2L{y}sy(0)y(0))+28L(sL{y}y(0))12L{y}=((s+2)(s2)(s2+4)216(2s+3)3)
Step 2
Plug the initial conditions , y(0)=0,y(0)=1,y0)=0,y(0)=1
6(s4L{y}s3y(0)s2y(0)sy"(0)y(0))+7(s3L{y}s2y(0)sy(0)y"(0))+21(s2L{y}sy(0)y(0))+28(sL{y}y(0))12L{y}=((s+2)(s2)(s2+4)216(2s+3)3)
6(s4L{y}s3×0s2×1s×0(1))+7(s3L{y}s2×0s×1(1))+21(s2L{y}s×01)+28(sL{y}0)12L{y}=
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