# How to solve this equation y'''-8y"+4y'+48= 4x^2+1 using Laplace Transformation.

How to solve this equation ${y}^{‴}-8y"+4{y}^{\prime }+48=4{x}^{2}+1$ using Laplace Transformation.
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Step 1
Converting the equation in Laplace transform:
${s}^{3}Y-8{s}^{2}Y+4sY+48Y=4\frac{{d}^{2}}{d{s}^{2}}\frac{1}{s}+\frac{1}{s}$
${s}^{3}Y-8{s}^{2}Y+4sY+48Y=\frac{8}{{s}^{3}}+\frac{1}{s}$
${s}^{3}Y-8{s}^{2}Y+4sY+48Y=\frac{\left(8+{s}^{2}\right)}{\left({s}^{3}\right)}$
$\left({s}^{3}-8{s}^{2}+4s+48\right)Y=\frac{\left(8+{s}^{2}\right)}{\left({s}^{3}\right)}$
$Y=\frac{\left(8+{s}^{2}\right)}{{s}^{3}\left({s}^{3}-8{s}^{2}+4s+48\right)}$
Step 2
Converting into partial fraction and solving:
$Y=\frac{\left(8+{s}^{2}\right)}{{s}^{3}\left({s}^{3}-8{s}^{2}+4s+48\right)}$
$=\frac{\left(8+{s}^{2}\right)}{{s}^{3}\left(s-4\right)\left(s+2\right)\left(s-6\right)}$
$=\frac{1}{\left(6{s}^{3}\right)}-\frac{1}{\left(72{s}^{2}\right)}-\frac{1}{32\left(s-4\right)}+\frac{43}{864s}-\frac{1}{32\left(x+2\right)}+\frac{11}{864\left(s-6\right)}$
$=\frac{1}{12}\frac{{d}^{2}}{d{s}^{2}}\frac{1}{s}+\frac{-1}{72}\frac{d}{ds}\frac{1}{s}-\frac{1}{32\left(s-4\right)}+\frac{43}{864s}-\frac{1}{32\left(s+2\right)}+\frac{11}{864\left(s-6\right)}$
Step 3
Thus finding inverse Laplace transform find y(t):
$y\left(t\right)=\frac{\left({x}^{2}\right)}{12}+\frac{x}{72}-\frac{{e}^{4x}}{32}+\frac{43}{864}-\frac{{e}^{-2x}}{32}+\frac{11{e}^{6x}}{864}$