# If x''+x'-6x=e^(-t), with x(0)=x'(0)=0, what is the value of X(s)?

If , what is the value of X(s)?

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Layton
Step 1
Given

To find X(s)
Step 2
Laplace transformation
$L\left\{x{}^{″}\right\}={s}^{2}X\left(s\right)-sx\left(0\right)-{x}^{\prime }\left(0\right)$
$L\left\{{x}^{\prime }\right\}=sX\left(s\right)-x\left(0\right)$
$L\left\{x\right\}=X\left(s\right)$
$L\left\{{e}^{-t}\right\}=\frac{1}{s+1}$
Step 3
Laplace transformation of given differential equation is
$L\left\{x{}^{″}+{x}^{\prime }-6x\right\}=L\left\{{e}^{-t}\right\}$
${s}^{2}X\left(s\right)-sx\left(0\right)-{x}^{\prime }\left(0\right)+sX\left(s\right)-x\left(0\right)-6X\left(s\right)=\frac{1}{s+1}$
${s}^{2}X\left(s\right)+sX\left(s\right)-6X\left(s\right)=\frac{1}{s+1}$
$X\left(s\right)\left[{s}^{2}+s-6\right]=\frac{1}{s+1}$
$X\left(s\right)\left[\left(s+3\right)\left(s-2\right)\right]=\frac{1}{s+1}$
$X\left(s\right)=\frac{1}{\left(s-2\right)\left(s+1\right)\left(s+3\right)}$