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Chapter 7: Q10E (page 404)

Use the convolution theorem to find the inverse Laplace transform of the given function.

1s3(s2+1)

Short Answer

Expert verified

The inverse Laplace transform for the given function by using the convolution theorem is.

y(t)=12t2-2+2cost

Step by step solution

01

 Step 1: Define convolution theorem

Let f(t)and g(t)be piecewise continuous on [0,鈭�)and of exponential order and set F(s)=L{f}(s)andG(s)=L{g}(s), then,

L{f*g}(s)=F(s)G(s),
or

L-1{F(s)G(s)}(t)=(f*g)(t)

02

Find the inverse Laplace transform for the given function

Consider the function,

1s3s2+1

Let, y(s)=1s3s2+1

Take inverse Laplace transform on both sides,

L-1[y(s)]=L-11s3s2+1

Use the convolution formula, L-1[f(s)路g(s)]=f*g=鈭�0tf(t-v)g(v)dv, where

f(s)=1s2+1andf(t)=sint

g(s)=1s3andg(t)=12t2

Hence, the equation 1 becomes,

y(t)=鈭�0tsin(t-v)路12v2dv

Thus, by using parts rule,

role="math" localid="1664877645923" y(t)=12v2cos(t-v)-鈭�0t2vcos(t-v)dv0t=12v2cos(t-v)-2[-vsin(t-v)+cos(t-v)]0t=12t2cos0-2[-tsin0+cos0-0+2[0+cost]]=12t2-2+2cost

Therefore, the inverse Laplace transform for the given function is.

y(t)=12t2-2+2cost

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Most popular questions from this chapter

solve the given initial value problem using the method of Laplace transforms. Sketch the graph of the solution.

y''+y=u(t--3);y(0)=0,y'(0)=1

In Problems 1鈥�19, use the method of Laplace transforms to solve the given initial value problem. Here x鈥�, y鈥�, etc., denotes differentiation with respect to t; so does the symbol D.

x'+y=1鈭�u(t鈭�2);x(0)=0,x+y'=0;y(0)=0

In Problems 1-14 , solve the given initial value problem using the method of Laplace transforms.

y''+6y'+5y=12et;y0=-1,y'0=7

In Problems 15-24, solve for Ys , the Laplace transform of the solution ytto the given initial value problem.

y''-3y'+2y=cost;y0=0,y'0=-1

In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.

y''+5y'+6y=g(t):y(0)=0,y'(0)=2,whereg(t)={0,0猢�t<1,t,1<t<5,1,5<t
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