 # 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 alesterp 2021-03-07 Answered

use the Laplace transform to solve the initial value problem.
$y"-3{y}^{\prime }+2y=\left\{\begin{array}{ll}0& 0\le t<1\\ 1& 1\le t<2\\ -1& t\ge 2\end{array}$
$y\left(0\right)=-3$
${y}^{\prime }\left(0\right)=1$

You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it pierretteA

Step 1
It is required to solve the initial value problem using laplace transform:
$y"-3{y}^{\prime }+2y=\left\{\begin{array}{ll}0& 0\le t<1\\ 1& 1\le t<2\\ -1& t\ge 2\end{array}$
$y\left(0\right)=-3$
${y}^{\prime }\left(0\right)=1$
Step 2
The first step in using laplace transform to solve an initial value problem is to take the transform of every term of differential equation:
$L\left(y"\right)-3L\left({y}^{\prime }\right)+2L\left(y\right)=0,0\le t<1$
Step 3
Now, use the appropriate formula’s:
${s}^{2}Y-sy\left(0\right)-{y}^{\prime }\left(0\right)-3\left(sY\left(s\right)-y\left(0\right)\right)+2\left(Y\left(s\right)\right)=L\left(0\right),0\le t<1$
${s}^{2}Y\left(s\right)-s\left(-3\right)-\left(1\right)-3sY\left(s\right)+3\left(-3\right)+2Y\left(s\right)=0$
$\left({s}^{2}-3s+2\right)Y\left(s\right)=10-3s$
$Y\left(s\right)=\frac{10-3s}{\left({s}^{2}-3s+2\right)}⇒Y\left(s\right)=\frac{10-3s}{\left(s-1\right)\left(s-2\right)}$
Step 4
Now, find the partial fraction decomposition for the transform:
$\frac{10-3s}{\left(s-1\right)\left(s-2\right)}=\frac{A}{\left(s-1\right)}+\frac{B}{\left(s-2\right)}$
$=\frac{A\left(s-2\right)+B\left(s-1\right)}{\left(s-1\right)\left(s-2\right)}$
$10-3s=As-2A+Bs-B$
$10-3s=\left(A+B\right)s+\left(-2A-B\right)$
$A+B=-3$
$-2A-B=10$
after solving : A=-7 and B=4
Step 5
Thus,
$Y\left(s\right)=\frac{10-3s}{\left(s-1\right)\left(s-2\right)}=\frac{A}{\left(s-1\right)}+\frac{B}{\left(s-2\right)}=\frac{-7}{\left(s-1\right)}+\frac{4}{\left(s-2\right)}$
finally taking the inverse transform:
$y\left(t\right)=-7{e}^{t}+4{e}^{2t}$
Step 6
Now, consider the case when $f\left(t\right)=1,1\le t<2$
${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)-3\left(sY\left(s\right)-y\left(0\right)\right)+2\left(Y\left(s\right)\right)=L\left(1\right),1\le t<2$
${s}^{2}Y\left(s\right)-s\left(-3\right)-\left(1\right)-3sY\left(s\right)+3\left(-3\right)+2Y\left(s\right)=\frac{1}{s}$
$\left({s}^{2}-3s+2\right)Y\left(s\right)=10-3s+\frac{1}{s}$
$Y\left(s\right)=\frac{10-3s+\frac{1}{s}}{{s}^{2}-3s+2}⇒Y\left(s\right)=\frac{-3{s}^{2}+10s+1}{s\left(s-1\right)\left(s-2\right)}$
Step 7
Now, find the partial fraction decomposition for the transform:
$\frac{-3{s}^{2}+10s+1}{s\left(s-1\right)\left(s-2\right)}=\frac{A}{s}+\frac{B}{s-1}+\frac{C}{s-2}$