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

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

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