Laplace transform

Difterential equation,

$y^{\text{'}\text{'}}+16y=4e^{4t},y(0)=2,y^{\text{'}}(0)=0$

We know that

$L\{y\}=Y(s)$

$L\{y^{\text{'}\text{'}}\}=s^{2}Y(s)-sy(0)-y^{\text{'}}(0)$

$L\{e^{at}\}=\frac{1}{s-a}$

Now taking Laplace transform of given differential equation

$L\{y^{\text{'}\text{'}}+16y=4e^{4t}\}$

$s^{2}Y(s)-sy(0)-y^{\text{'}}(0)+16Y(s)=\frac{4}{s-4}$

$(s^2+16)Y(s)-s(2)-(0)=\frac{4}{s-4}$

$(s^2+16)Y(s)=\frac{4}{s-4}+2s$

$Y(s)=\frac{4}{(s-4)(s^2+16)}+\frac{2s}{s^2+16}$

$Y(s)=\frac{4}{(s-4)(s^2+16)}+\frac{2s}{s^2+16}$

$Y(s)=\frac{1}{8(s-4)}+\frac{15}{8}\frac{s}{s^2+4^2}-\frac{1}{8}\frac{4}{s^2+4^2}$

Help me how to get $1/8($s$-4)+15/8$

s$/$s$^2+4^2-1/84/$s$^2+4^2$