In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative frac{dy}{dt} also appears. Consid

Braxton Pugh 2021-02-09 Answered
In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative dydt also appears. Consider the following initial value problem, defined for t > 0:
dydt+40ty(tw)e4wdw=3,y(0)=0
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
Y(s)=L{y(t))}?
b) Obtain the solution y(t).
y(t) - ?
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Expert Answer

FieniChoonin
Answered 2021-02-10 Author has 102 answers

Step 1
Given,  dy  dt +40ty(tw)e4wdw=3,y(0)=0
a) Taking Laplace transform of the above equation
L(y)+4L(0ty(tw)e4wdw)=L(3)
sy(s)y(0)+4y(s)s+4=3s
y(s)[s+4s+4]=3s
y(s)[s2+4s+4s+4]=3s
y(s)=3s[s+4s2+4s+4]
=3s[s+4(s+2)2]
y(s)=3(s+4)s(s+2)2
Step 2
Now take, 3(s+4)s(s+2)2 resolve into partial fractions
3(s+4)s(s+2)2=As+Bs+2+C(s+2)2
3(s+4)s(s+2)2=A(s+2)2+Bs(s+2)+Css(s+2)2
3(s+4)=A(s+2)2+Bs(s+2)+Cs
Put s=0
12=4A+0+0
A=3
Put s=2
6=0+02C
C=3
Consider coefficients of 's2'
0=A+B
A=B or B=A
B=3
Put the values in the equation
3(s+4)s(s+2)2=3s+3s+2+3(s+2)2
 

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