Answer to Question #324463 in Differential Equations for Genius

Question

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

y''(t) + 2y'(t) -3y(t) = e^-3t if y'(0) = 0 and y(0) = 0

Answer 1

$y''\left( t \right) +2y'\left( t \right) -3y\left( t \right) =e^{-3t},y'\left( 0 \right) =0,y\left( 0 \right) =0\\y\left( t \right) \rightarrow Y\left( p \right) \\p^2Y\left( p \right) -py\left( 0 \right) -y'\left( 0 \right) +2\left( pY\left( p \right) -y\left( 0 \right) \right) -3Y\left( p \right) =\frac{1}{p+3}\\\left( p^2+2p-3 \right) Y\left( p \right) =\frac{1}{p+3}\\Y\left( p \right) =\frac{1}{\left( p+3 \right) ^2\left( p-1 \right)}\\\frac{1}{\left( p+3 \right) ^2\left( p-1 \right)}=\frac{A}{\left( p+3 \right) ^2}+\frac{B}{p+3}+\frac{C}{p-1}\\1=A\left( p-1 \right) +B\left( p+3 \right) \left( p-1 \right) +C\left( p+3 \right) ^2\\\left( B+C \right) p^2+\left( A+2B+6C \right) p+\left( -A-3B+9C \right) =1\\\left\{ \begin{array}{c} B+C=0\\ A+2B+6C=0\\ -A-3B+9C=1\\\end{array} \right. \Rightarrow A=-\frac{1}{4},B=-\frac{1}{16},C=\frac{1}{16}\\Y\left( p \right) =-\frac{1}{4\left( p+3 \right) ^2}-\frac{1}{16\left( p+3 \right)}+\frac{1}{16\left( p-1 \right)}\mapsto -\frac{1}{4}te^{-3t}-\frac{1}{16}e^{-3t}+\frac{1}{16}e^t\\y\left( t \right) =-\frac{1}{4}te^{-3t}-\frac{1}{16}e^{-3t}+\frac{1}{16}e^t$