Question

Find the general solution.

1. 4y^(4) −4y′′′ −23y′′ +12y′ +36y = 0 

Answer 1

$4y^{\left( 4 \right)}-4y'''-23y''+12y'+36y=0\\4\lambda ^4-4\lambda ^3-23\lambda ^2+12\lambda +36=0\\\left( \lambda -2 \right) ^2\left( 2\lambda +3 \right) ^2=0\\\lambda _{1,2}=2,\lambda _{3,4}=-\frac{3}{2}\\y=C_1e^{2t}+C_2te^{2t}+C_3e^{-\frac{3}{2}t}+C_4te^{-\frac{3}{2}t}$