Solve the following initial value problems using Laplace Transforms: a) frac{d^2y}{dt^2}+4frac{dy}{dt}+3y=1, y(0=0) , y'(0)=0 b) frac{d^2y}{dt^2}+4frac{dy}{dt}=1, y(0=0) , y'(0)=0 c) 2frac{d^2y}{dt^2}+3frac{dy}{dt}-2y=te^{-2t}, y(0=0) , y'(0)=-2

sanuluy 2021-01-10 Answered
Solve the following initial value problems using Laplace Transforms:
a)d2ydt2+4dydt+3y=1,
y(0=0),y(0)=0
b)d2ydt2+4dydt=1,
y(0=0),y(0)=0
c)2d2ydt2+3dydt2y=te2t,
y(0=0),y(0)=2
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Expert Answer

Fatema Sutton
Answered 2021-01-11 Author has 88 answers

Step 1 We need to solve the given differential equations using Laplace transformation.
a)d2ydt2+4dydt+3y=1,
y(0=0),y(0)=0
Now , taking Laplace transformation both of the sides.
L{d2ydt2+4dydt+3y}=L{1}
s2Y(s)sy(0)y(0)+4(sY(s)y(0))+3Y(s)=1s
(s2+4s+3)Y(s)=1s( Using y(0)=0,y(0)=0)
Y(s)=1s2+4s+3
Y(s)=1s(s+1)(s+3)
Y(s)=13s12(s+1)+16(s+3) (By the method of partial fraction)
Step 2
Now, taking inverse Laplace transformation to get the final answer.
y(t)=L1{13s12(s+1)+16(s+3)}
y(t)=13L1{1s}12L1{1s+1}+16L1{1s+3}
y(t)=13112et+16e3t
y(t)=13et2+e3t6
Answer y(t)=13et2+e3t6
Step 3
b)d2ydt2+4dydt=1,
y(0=0),y(0)=0
Now , taking Laplace transformation both of the sides. L{d2ydt2+4dydt}=L{1}
s2Y(s)sy(0)y(0)+4(sY(s)y(0))=1s
(s2+4s)Y(s)=1s(Using y(0)=0,y(0)=0)
Y(s)=1s(s2+4s)
Y(s)=1s(s+1)(s+3)
Y(s)=14s2+116(s+4)116s (By the method of partial fraction)
Step 4
Now, taking inverse Laplace transformation to get the final answer.

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