Use Laplace transforms to solve the following initial value problem y"-y'-6y=0 y(0)=1 y'(0)=-1

Step 1
The given IVP is as follows.
$y"-y^{\prime }-6y=0$
$y(0)=1$
$y^{\prime }(0)=-1$
Apply Laplace transform on the IVP as follows.
$y^{″}-y^{\prime }-6y=0$
${\displaystyle L\left\{y{}^{″}\right\}-L\left\{y^{\prime }\right\}-L\left\{6y\right\}=L\left\{0\right\}}$
${\displaystyle s^{2}L\left\{y\right\}-sy\left(0\right)-y^{\prime }\left(0\right)-sL\left\{y\right\}+y\left(0\right)-6L\left\{y\right\}=0}$
${\displaystyle L\left\{y\right\}[s^2-s-6]=s\left(1\right)+(-1)-\left(1\right)}$
${\displaystyle L\left\{y\right\}[s^2-s-6]=s-2}$
${\displaystyle L\left\{y\right\}=\frac{s-2}{s^2-s-6}}$
${\displaystyle L\left\{y\right\}=\frac{s-2}{(s-3)(s+2)}}$
Step 2
Apply partial fraction on ${\displaystyle \frac{s-2}{(s-3)(s+2)}}$ as follows.
${\displaystyle \frac{s-2}{(s-3)(s+2)}=\frac{A}{s-3}+\frac{B}{s+2}}$
$s-2=A(s+2)+B(s-3)$
$s-2=(A+B)s+(2A-3B)$
$A+B=1,2A-3B=-2$
$A=1-B$
$2(1-B)-3B=-2$
$-5B=-4$
${\displaystyle B=\frac{4}{5}}$
${\displaystyle A=\frac{1}{5}}$
${\displaystyle \frac{s-2}{(s-3)(s+2)}=\frac{1}{5}\left(\frac{1}{s-3}\right)+\frac{4}{5}\left(\frac{1}{s+2}\right)}$
Obtain the solution of the IVP as follows.
${\displaystyle L\left\{y\right\}=15(1s-3)+45(1s+2)}$
Take inverse Laplace transforms as follows.
${\displaystyle L^{-1}\left\{L\left\{y\right\}\right\}=L^{-1}\{\frac{1}{5}\left(\frac{1}{s-3}\right)+\frac{4}{5}\left(\frac{1}{s+2}\right\}}$

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