Question

Determine the unique solution of the following differential equations by using Laplace transforms:

y''(t)-6y'(t)+9y(t) = t²e^3t if y'(0) = 6 and y(0) = 2

Answer 1

$y''-6y'+9y=t^2e^{3t},y'\left( 0 \right) =6,y\left( 0 \right) =2\\y\left( t \right) \rightarrow Y\left( p \right) \\p^2Y\left( p \right) -py\left( 0 \right) -y'\left( 0 \right) -6\left( pY\left( p \right) -y\left( 0 \right) \right) +9Y\left( p \right) =\frac{2!}{\left( p-3 \right) ^3}\\p^2Y\left( p \right) -2p-6-6\left( pY\left( p \right) -2 \right) +9Y\left( p \right) =\frac{2}{\left( p-3 \right) ^3}\\Y\left( p \right) =\frac{2}{\left( p-3 \right) ^5}+\frac{2}{p-3}\rightarrow 2\cdot \frac{t^4}{4!}e^{3t}+2e^{3t}=\\=\frac{1}{12}t^4e^{3t}+2e^{3t}$