 # Solve the question with laplace transformation y''+2y'+y=4 y(0)=3 y'(0)=0 he298c 2021-02-08 Answered
Solve the question with laplace transformation
${y}^{″}+2{y}^{\prime }+y=4$
$y\left(0\right)=3$
${y}^{\prime }\left(0\right)=0$
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To solve the differential equation with Laplace Transformation
${y}^{″}+2{y}^{\prime }+y=4$
$y\left(0\right)=3$
${y}^{\prime }\left(0\right)=0$
Step 2
Laplace transformation of y'' is
$L\left\{{y}^{″}\right\}={s}^{2}Y-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left\{{y}^{″}\right\}={s}^{2}Y-3s$
Laplace transformation of y' is
$L\left\{{y}^{\prime }\right\}=sY-y\left(0\right)$
$L\left\{{y}^{\prime }\right\}=sY-3$
Laplace transformation of y is
$L\left\{y\right\}=Y$
Laplace transformation of 4 is
$L\left\{4\right\}=\frac{4}{s}$
Step 3
Laplace transformation of given differential equation
$L\left\{{y}^{″}+2{y}^{\prime }+y\right\}=L\left\{4\right\}$
${s}^{2}Y-3s+2\left(sY-3\right)+Y=\frac{4}{s}$
${s}^{2}Y+2sY+Y-3s-6=\frac{4}{s}$
$\left({s}^{2}+2s+1\right)Y=\frac{4}{s}+3s+6$
$\left(s+1{\right)}^{2}Y=\frac{\left(3{s}^{2}+6s+4\right)}{s}$
$Y=\frac{\left(3{s}^{2}+6s+3+1\right)}{s\left(s+1{\right)}^{2}}$
$Y=\frac{3\left({s}^{2}+2s+1\right)}{s\left(s+1{\right)}^{2}}+\frac{1}{s\left(s+1{\right)}^{2}}$
$Y=\frac{3\left(s+1{\right)}^{2}}{s\left(s+1{\right)}^{2}}+\frac{1}{s\left(s+1{\right)}^{2}}$
$Y=\frac{3}{s}+\frac{1}{s\left(s+1{\right)}^{2}}$
Step 4
$\frac{1}{s\left(s+1{\right)}^{2}}=\frac{A}{s}+\frac{B}{\left(s+1\right)}+\frac{C}{\left(s+1{\right)}^{2}}$
$1=A\left(s+1{\right)}^{2}+Bs\left(s+1\right)+\left(C\right)s$
$1=A{s}^{2}+2As+A+B{s}^{2}+Bs+Cs$
$1=\left(A+B\right){s}^{2}+\left(2A+B+C\right)s+A$
$A=1$
$A+B=0$
$1+B=0$
$B=-1$
$2A+B+C=0$
$2-1+C=0$
$C=-1$
Thus , $\frac{1}{s\left(s+1{\right)}^{2}}=\frac{1}{s}+\frac{1}{\left(s+1\right)}-\frac{1}{\left(s+1{\right)}^{2}}$
Step 5
Thus, from equation (1)
$Y=\frac{3}{s}+\frac{1}{s}-\frac{1}{\left(s+1\right)}-\frac{1}{\left(s+1{\right)}^{2}}$
$Y=\frac{4}{s}-\frac{1}{\left(s+1\right)}-\frac{1}{\left(s+1{\right)}^{2}}$
Step 6
Inverse Laplace transform of Y is
${L}^{-1}\left\{Y\right\}=y$
Inverse Laplace transform of $\frac{1}{s}$ is
${L}^{-1}\left\{\frac{1}{s}\right\}=1$
Inverse Laplace transform of $\frac{1}{\left(s+1\right)}$ is