Use Laplace Transform to solve the given equation: y'' + 6y' + 5y = t cdot U (t-2) y(0) = 1 y' (0) = 0

remolatg

remolatg

Answered question

2021-02-08

Use Laplace Transform to solve the given equation:
y+6y+5y=tU(t2)
y(0)=1
y(0)=0

Answer & Explanation

smallq9

smallq9

Skilled2021-02-09Added 106 answers

Step 1
Let Laplace transform of y(t) be Y(s). Then Laplace transform of y(t) and y"(t) is sY(s)y(0) or sY(s)1 and s2Y(s)sy(0)y(0) or s2Y(s)s
Also Laplace transform of u(t2) is e2ss Therefore Laplace transform of tu(t2) is ddse2ss or (2s+1)e2ss2
Step 2
Finally substituting the results in the equation given:
s2Y(s)s+6(sY(s)1)+5Y(s)=(2s+1)e2ss2
(s2+6s+5)Y(s)s6=(2s+1)e2ss2
(s2+6s+5)Y(s)=(2s+1)e2ss2+s+6
Y(s)=(2s+1)e2ss2(s2+6s+5)+s+6s2+6s+5
Y(s)=(2s+1)e2ss2(s+5)(s+1)+s+6(s+5)(s+1)
Y(s)=e2s(3225s+75s2+54s+1+3100s+5)+54s+114s+5
Step 3
Now Laplace inverse of:
1s is u(t)
1s2 is tu(t)
1s+1 is etu(t)
1s+5 is e5tu(t)
Step 4
Thus using time shifting property and solving Y(s)=e2s(3225s+75s2+54s+1+3100s+5)+54s+114s+5:
y(t)=(3225u(t2))+(75(t2)u(t2))+(54e2tu(t2))+(3100e105tu(t2))+(54etu(t))(14e5tu(t))

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