Question

Use the Laplace transform to solve the following IVP

y′′ +2y′ +2y=δ(t−5) ,y(0)=1,y′(0)=2,

where δ(t) is the Dirac delta function.

Answer #1

Taking the Laplace on the LHS,

We know,

So the equation becomes,

Substituting the initial values,

y'(0)=2 and y(0)=1,

Laplace of RHS,

Now,

So,

By splitting,

where, H is the Heaviside step function and

So,

thus,

Use
Laplace transform to solve IVP
2y”+2y’+y=2t , y(0)=1 , y’(0)=-1

Use the Laplace transform to solve the IVP:
y′(t) +y(t) = cos(t),
y(0) = 0.

Use laplace transform to solve the given IVP
y''-2y'-48y=0
y(0)=13
y'(0)=6

Take the Laplace transform of the following initial value
problem and solve for Y(s)=L{y(t)}: y′′−2y′−35y=S(t)y(0)=0,y′(0)=0
where S is a periodic function defined by S(t)={1,0≤t<1 0,
1≤t<2, and S(t+2)=S(t) for all t≥0. Hint: : Use the formula for
the Laplace transform of a periodic function.
Y(s)=

Use the definition of the Laplace transform to solve the
IVP:
4y''− 4y' + 5y = δ(t), y(0) = −1, y'(0) = 0.

Given the differential equation
y''−2y'+y=0, y(0)=1, y'(0)=2
Apply the Laplace Transform and solve for Y(s)=L{y}
Y(s) =
Now solve the IVP by using the inverse Laplace Transform
y(t)=L^−1{Y(s)}
y(t) =

Use the Laplace transform to solve the following initial value
problem:
y′′−4y′−32y=δ(t−6)y(0)=0,y′(0)=0

Use the Laplace transform to solve the given initial-value
problem. y'' + y = δ(t − 8π), y(0) = 0, y'(0) = 1

Use Laplace transform to solve the following initial value
problem: y '' − 2y '+ 2y = e −t , y(0) = 0 and y ' (0) =
1
differential eq

solve using the laplace transform y''-2y'+y=e^-t , y(0)=0 ,
y'(0)=1

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