# Solve by using Laplace Transform and explain the steps in brief displaystylefrac{{{d}^{2}{x}}}{{{left.{d}{t}right.}^{2}}}+{2}frac{{{left.{d}{x}right.}}}{{{left.{d}{t}right.}}}+{x}={3}{t}{e}^{{-{t}}} Given x(0)=4, x'(0)=2

Solve by using Laplace Transform and explain the steps in brief
$\frac{{d}^{2}x}{{dt}^{2}}+2\frac{dx}{dt}+x=3t{e}^{-t}$
Given $x\left(0\right)=4,{x}^{\prime }\left(0\right)=2$
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Step 1
The given differential equation is:
$\frac{{d}^{2}x}{{dt}^{2}}+2\frac{dx}{dt}+x=3t{e}^{-t}\dots \left(1\right)$
Given: $x\left(0\right)=4,{x}^{\prime }\left(0\right)=2$
Step 2
Taking Laplace transform on both sides of (1). we have,
$L\left\{\frac{{d}^{2}x}{{dt}^{2}}\right\}+2L\left\{\frac{dx}{dt}\right\}+L\left\{x\right\}=3L\left\{t{e}^{-t}\right\}$
$⇒{x}^{2}L\left(x\right)-x\cdot x\left(0\right)-{x}^{\prime }\left(0\right)+2L\left(x\right)-2x\left(0\right)+L\left(x\right)=\frac{3}{{\left(x+1\right)}^{2}}$
$⇒{x}^{2}L\left(x\right)-4x-2+2\left[L\left(x\right)-4\right]+L\left(x\right)=\frac{3}{{\left(x+1\right)}^{2}}$
$⇒L\left(x\right)\left\{{x}^{2}+2x+1\right\}=\frac{3}{{\left(x+1\right)}^{2}}+4x+10$
$⇒L\left(x\right)=\frac{3}{{\left(x+1\right)}^{4}}+\frac{4x+10}{{\left(x+1\right)}^{2}}$
Step 3
Now by taking inverse Laplace transformation,
${L}^{-1}\left\{L\left(x\right)\right\}=3{L}^{-1}\left\{\frac{1}{{\left(x+1\right)}^{4}}\right\}+{L}^{-1}\left\{\frac{4x+10}{{\left(x+1\right)}^{2}}\right\}$
${L}^{-1}\left\{L\left(x\right)\right\}=3{L}^{-1}\left\{\frac{1}{{\left(x+1\right)}^{4}}\right\}+{L}^{-1}\left\{\frac{4x+10}{{\left(x+1\right)}^{2}}\right\}$
$⇒x=3{L}^{-1}\left\{\frac{1}{{\left(x+1\right)}^{4}}\right\}+{L}^{-1}\left\{\frac{4x+10}{{\left(x+1\right)}^{2}}\right\}$