Use the definition of Laplace Transforms to show that:displaystyle{L}{leftlbrace{t}^{n}rightrbrace}=frac{{{n}!}}{{{s}^{{{n}+{1}}}}},{n}={1},{2},{3},ldots

Jason Farmer 2021-02-12 Answered

Use the definition of Laplace Transforms to show that:
L{tn}=n!sn+1,n=1,2,3,

You can still ask an expert for help

Want to know more about Laplace transform?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Nicole Conner
Answered 2021-02-13 Author has 97 answers
Step 1
Given:
The laplace transformation,
L{tn}=n!sn+1
Step 2
Proof:
Consider L(tn) and simplify it as follows.
L{f(t)}=0estf(t)dt
L{tn}=0est(tn)dt
Apply integration by parts, uv=uvintuv
Here,
u=tndv=est
du=ntn+1   v=ests
L{tn}=[tnests]00ntn1(ests)dt
=0+ns0tn1(est)dt
=nsL(tn1)
Step 3
Now substitute the values 1, 2, 3... for n.
When n=1
L{t1}=1sL(1)
=1s(1s)
=1s2
When n=2
L{t2}=2s(L(t1))
=2s(1s2)
=2!s3
Step 4
When n = 3,
L{t3}=3sL(t31)
=3s(L(t2))
=3s(2s3)
=6s3+1
=3!s3+1
Step 5
Thus, for n = n,
L(tn)=n!sn+1
Hence it is proved.

We have step-by-step solutions for your answer!

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2021-09-03

Solve the following differential equations using the Laplace transform
y"4y=0
y(0)=0,y(0)=0

asked 2022-01-18

Obtaining Differential Equations from Functions
dydx=x21
is a first order ODE,
d2ydx2+2(dydx)2+y=0
is a second order ODE and so on. I am having trouble to obtain a differential equation from a given function. I could find the differential equation for
y=ex(Acosx+Bsinx) and the steps that I followed are as follows.
dydx=ex(Acosx+Bsinx)+ex(Asinx+Bcosx)
=y+ex(Asinx+Bcosx) (1)
d2ydx2=dydx+ex(Asinx+Bcosx)+ex(AcosxBsinx)
=dydx+(dydxy)y
using the orginal function and (1). Finally,
d2ydx22dydx+2y=0,
which is the required differential equation.
Similarly, if the function is y=(Acos2t+Bsin2t), the differential equation that I get is
d2ydx2+4y=0
following similar steps as above.

asked 2021-10-23
Show that any separable equation M(x,y)+N(x,y)y=0
asked 2022-01-20
What might be a solution to the differential equation of the form
xy=cyy+d
where y=y(x) and c,d are constants? I am supposed to simply state a solution to this, but I don;t think it is all that obvious.
asked 2022-01-21
A simple question about the solution of homogeneous equation to this differential equation
Given that t,1+t,t2,t are the solutions to y+a(t)y+b(t)y+c(t)y=d(t), what is the solution of homogeneous equation to this differential equation? What i have done is tried the properties of linear differential equation that
L(t)=L(1+t)=L(t2)=L(t)=d(t) so the homogeneous solution should be independent and i claim that 1,t,t2 should be the solution. However, i am not sure hot can i actually conclude that these are the solutions? It seems that it can be quite a number of sets of solution by the linearity.
asked 2022-01-15
Solve the differential equation:
y(x)=y(x)x1+1xy(x)+2
with initial condition y(0)=0
asked 2021-06-16
2(dydx)+2y=0

New questions