By using Laplace transforms, solve the following differential equations subject to the given initial conditions. y"-4y'=-4te^{2t}, y_0=0, y'_0 =1 y"+2y'+10y=-6e^{-t}sin3t, y_0=0, y'_0=1

he298c 2021-02-15 Answered

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. 
y"-4y'=-4te2t,y0=0,y'0=1

y"+2y'+10y=-6e-t sin3t,y0=0,y'0=1

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Expert Answer

firmablogF
Answered 2021-02-16 Author has 92 answers

Step 1The given differential equation is,
y"4y=4te2t
where y0=0,y0=1
Take the Laplace Transform on both sides,
L(y4y)=L(4te2t)
L(y)4L(y)=4L(te2t)
Consider the formulae,
L(y)=s2L(y)sy0y0
L(y)=sL(y)y0
L(te2t)=1(s2)2
Step 2
Substitute the values,
L(y)4L(y)=4L(te2t)
s2L(y)sy0y04[sL(y)y0]=41(s2)2
s2L(y)s(0)(1)4sL(y)=(4)(s2)2
L(y)[s24s]=1+(4)(s2)2
L(y)=s24s+44(s2)2(s24s)
L(y)=1(s2)2
Step 3
Take the Inverse Laplace Transform both sides,
L1[L(y)]=L1[1(s2)2]
y=L1[1(s2)2]
y=te2t
Thus, the solution of the given integral is y=te2tii)
The given differential equation is,
y+24y+10y=6etsin3t
where y0=0,y0=1
Step 4
Take the Laplace Transform on both sides,
L(y+24y+10y)=L[6etsin(3t)]
L(y)+24L(y)+10L(y)=6L[etsin(3t)]
Use the same formulae as mentioned above,
s2L(y)01+2[sL(y)0]+10L(y)=(6)3(s+1)2+9
s2L(y)1+2sL(y)+10L(y)=18s2+2s+1+9
[s2+2s+10]L(y)=118s2+2s+10
L(y)=s2+2s+1018(s2+2s+10)2
L(y)=s2+2s-8(s2+2s+10)2

Step 5

Take the Inverse Laplace Transform both sides, 
L-1[L(y)]=L-1s2+2s-8(s2+2s+10)2

y=te-tcos(3t)

Thus, the solution of the given differential equation is y=te-tcos(3t)

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