use the Laplace transform to solve the initial value problem.y"−3y′+2y={00≤t&lt;111≤t&lt;2−1t≥2y(0)=−3y′(0)=1

Name: use the Laplace transform to solve the initial value problem.y&quot;−3y′+2y={00≤t&amp;lt;111≤t&amp;lt;2−1t≥2y(0)=−3y′(0)=1
Uploaded: 2022-02-23T17:22:57+00:00
Description: use the Laplace transform to solve the initial value problem.y&quot;−3y′+2y={00≤t&amp;lt;111≤t&amp;lt;2−1t≥2y(0)=−3y′(0)=1

Question

use the Laplace transform to solve the initial value problem.y&quot;−3y′+2y={00≤t&amp;lt;111≤t&amp;lt;2−1t≥2y(0)=−3y′(0)=1

pierretteA · Accepted Answer

Step 1It is required to solve the initial value problem using laplace transform:y&quot;−3y′+2y={00≤t&amp;lt;111≤t&amp;lt;2−1t≥2y(0)=−3y′(0)=1Step 2The first step in using laplace transform to solve an initial value problem is to take the transform of every term of differential equation:L(y&quot;)−3L(y′)+2L(y)=0,0≤t&amp;lt;1Step 3Now, use the appropriate formula’s:s2Y−sy(0)−y′(0)−3(sY(s)−y(0))+2(Y(s))=L(0),0≤t&amp;lt;1s2Y(s)−s(−3)−(1)−3sY(s)+3(−3)+2Y(s)=0(s2−3s+2)Y(s)=10−3sY(s)=10−3s(s2−3s+2)⇒Y(s)=10−3s(s−1)(s−2)Step 4Now, find the partial fraction decomposition for the transform:10−3s(s−1)(s−2)=A(s−1)+B(s−2)=A(s−2)+B(s−1)(s−1)(s−2)10−3s=As−2A+Bs−B10−3s=(A+B)s+(−2A−B)A+B=−3−2A−B=10after solving : A=-7 and B=4Step 5Thus,Y(s)=10−3s(s−1)(s−2)=A(s−1)+B(s−2)=−7(s−1)+4(s−2)finally taking the inverse transform:y(t)=−7et+4e2tStep 6Now, consider the case when f(t)=1,1≤t&amp;lt;2s2Y(s)−sy(0)−y′(0)−3(sY(s)−y(0))+2(Y(s))=L(1),1≤t&amp;lt;2s2Y(s)−s(−3)−(1)−3sY(s)+3(−3)+2Y(s)=1s(s2−3s+2)Y(s)=10−3s+1sY(s)=10−3s+1ss2−3s+2⇒Y(s)=−3s2+10s+1s(s−1)(s−2)Step 7Now, find the partial fraction decomposition for the transform:−3s2+10s+1s(s−1)(s−2)=As+Bs−1+Cs−2

use the Laplace transform to solve the initial value problem.y"-3y'+2y=begin{cases}0&0leq t<11&1leq t<2 -1&tgeq2end{cases}y(0)=-3y'(0)=1

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