 # Solve by using Laplace transform y"+y=cos t ,y(0)=0 , y'(0)=-1 Brittney Lord 2020-10-28 Answered
Solve by using Laplace transform
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Step 1
The given initial value problem is
Apply Laplace transform on both sides of the given equation as follows.
$L\left\{y"\right\}+L\left\{y\right\}=L\left\{\mathrm{cos}t\right\}$
${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+Y\left(s\right)=\frac{s}{{s}^{2}+1}$
$Y\left(s\right)\left({s}^{2}+1\right)+1=\frac{s}{{s}^{2}+1}$
$Y\left(s\right)=\frac{s-{s}^{2}-1}{\left({s}^{2}+1{\right)}^{2}}$
$\text{Step 2}$

$\frac{s-{s}^{2}-1}{\left({s}^{2}+1{\right)}^{2}}=\frac{As+B}{\left({s}^{2}+1\right)}+\frac{Cs+D}{\left({s}^{2}+1{\right)}^{2}}$
$=\frac{\left(As+B\right)\left({s}^{2}+1\right)+Cs+D}{\left({s}^{2}+1{\right)}^{2}}$

$s-{s}^{2}-1=\left(As+B\right)\left({s}^{2}+1\right)+Cs+D$
$=A{s}^{3}+As+B{s}^{2}+B+Cs+D$
$=A{s}^{3}+B{s}^{2}+\left(A+C\right)s+\left(B+D\right)$
$\text{Step 3}$

$A=0$
$B=-1$
$A+C=1⇒0+C=1⇒C=1$
$B+D=-1⇒D=-1+1⇒D=0$

$\frac{s-{s}^{2}-1}{\left({s}^{2}+1{\right)}^{2}}=\frac{As+B}{\left({s}^{2}+1\right)}+\frac{Cs+D}{\left({s}^{2}+1{\right)}^{2}}$
$=-\frac{1}{\left({s}^{2}+1\right)}+\frac{s}{\left({s}^{2}+1{\right)}^{2}}$
$\text{Step 4}$

$Y\left(s\right)=\frac{s-{s}^{2}-1}{\left({s}^{2}+1{\right)}^{2}}$
$=\frac{As+B}{\left({s}^{2}+1\right)}+\frac{Cs+D}{\left({s}^{2}+1{\right)}^{2}}$
$=-\frac{1}{\left({s}^{2}+1\right)}+\frac{s}{\left({s}^{2}+1{\right)}^{2}}$

${L}^{-1}\left\{Y\left(s\right)\right\}={L}^{-1}\left\{-\frac{1}{\left({s}^{2}+1\right)}+\frac{s}{\left({s}^{2}+1{\right)}^{2}}\right\}$