Solve the following differential equation by the laplace transform method. 6y^{(4)}+7y'''+21y"+28y'-12y=t(\cos(2t)+te^{\frac{-3t}{2}}).

Step 1
Given equation is
${\displaystyle 6y^{\left(4\right)}+7y{}^{‴}+21y28y^{\prime }-12y=t(\cos \left(2t\right)+t{e}^{\frac{-3t}{2}})}$
Apply Laplace transform on both sides
$L\{6y^{(4)}+7y^{‴}+21y"+28y^{\prime }-12y\}=L\{t(\cos (2t)+t{e}^{\frac{-3t}{2}})\}$
$6L\{y^{‴}(t)\}+7L\{y^{‴}(t)\}+21L\{y"(t)\}+28L\{y^{\prime }(t)\}+12L\left\{y\right\}=L\{t(\cos (2t)+t{e}^{\frac{-3t}{2}})\}$
$6(s^{4}L\left\{y\right\}-s^{3}y(0)-s^2{y}^{\prime }(0)-sy"(0)-y^{‴}(0))+7(s^{3}L\left\{y\right\}-s^{2}y(0)-s{y}^{\prime }(0)-y"(0))+21(s^{2}L\left\{y\right\}-sy(0)-y^{\prime }(0))+28L(sL\left\{y\right\}-y(0))-12L\left\{y\right\}=-(-\frac{(s+2)(s-2)}{(s^2+4{)}^2}-\frac{16}{(2s+3{)}^3})$
Step 2
Plug the initial conditions , ${\displaystyle y\left(0\right)=0,y^{\prime }\left(0\right)=1,y0)=0,y{}^{‴}\left(0\right)=-1}$
$6(s^{4}L\left\{y\right\}-s^{3}y(0)-s^2{y}^{\prime }(0)-sy"(0)-y^{‴}(0))+7(s^{3}L\left\{y\right\}-s^{2}y(0)-s{y}^{\prime }(0)-y"(0))+21(s^{2}L\left\{y\right\}-sy(0)-y^{\prime }(0))+28(sL\left\{y\right\}-y(0))-12L\left\{y\right\}=-(-\frac{(s+2)(s-2)}{(s^2+4{)}^2}-\frac{16}{(2s+3{)}^3})$