What is the Laplace inverse of s²+1/s²(s²-4s-9)

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Eliza Beth13 · Accepted Answer

To find the Laplace inverse of s2+1s2(s2−4s−9), we can use partial fraction decomposition and apply the inverse Laplace transform to each term.First, let&#039;s factor the denominator s2(s2−4s−9):s2(s2−4s−9)=s2(s−3)(s+3)We can express the given fraction as the sum of two fractions using partial fraction decomposition:s2+1s2(s2−4s−9)=As2+Bs+Cs+Ds2−4s−9To determine the values of A, B, C, and D, we can multiply both sides of the equation by the denominator and equate the numerators:s2+1=A(s2−4s−9)+B(s)(s2−4s−9)+(Cs+D)(s2)Expanding and collecting like terms, we get:s2+1=(A+C)s3+(−4A+B+D)s2+(−9A−4B+Cs)s−9ABy comparing coefficients, we obtain the following system of equations:A+C=0−4A+B+D=1−9A−4B+C=0−9A=1Solving this system of equations, we find A=−19, B=49, C=13, and D=179.Now, let&#039;s express the original fraction using the partial fraction decomposition:s2+1s2(s2−4s−9)=−19(1s2)+49(1s)+13s+179s2−4s−9We can now find the inverse Laplace transform of each term:ℒ−1{−19(1s2)}=−19tℒ−1{49(1s)}=49ℒ−1{13s+179s2−4s−9}=13ℒ−1{ss2−4s−9}+179ℒ−1{1s2−4s−9}For the term 13ℒ−1{ss2−4s−9}, we need to find the inverse Laplace transform of ss2−4s−9. We can complete the square in the denominator:s2−4s−9=(s−2)2−13This allows us to rewrite the fraction as:ss2−4s−9=s(s−2)2−13To find the inverse Laplace transform, we recognize that the expression is in the form of a shifted exponential function, specifically s(s−a)2+b2, which has the inverse Laplace transform 1beatsin(bt). In our case, a=2 and b=13. Therefore:ℒ−1{ss2−4s−9}=113e2tsin(13t)For the term 179ℒ−1{1s2−4s−9}, we recognize that it is the inverse Laplace transform of 1s2−4s−9. Using partial fraction decomposition, we can express it as:1s2−4s−9=As−3+Bs+3By equating the numerators, we have:1=A(s+3)+B(s−3)Solving for A and B, we find A=−16 and B=16. Thus:1s2−4s−9=−16(1s−3)+16(1s+3)The inverse Laplace transform of 1s2−4s−9 can be calculated as:ℒ−1{1s2−4s−9}=−16e3t+16e−3tPutting all the terms together, we have:ℒ−1{s2+1s2(s2−4s−9)}=−19t+49+13(113e2tsin(13t))+179(−16e3t+16e−3t)

What is the Laplace inverse of s²+1/s²(s²-4s-9)

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