# Question #ceca6

May 3, 2017

The characteristic equation is:

${D}^{3} + 3 {D}^{2} + 3 D + 1 = 0$

Or:

${\left(D + 1\right)}^{3} = 0 \implies D = - 1$

Repeated eigenvalues means that we have a complementary solution in the form:

${y}_{c} = {e}^{- x} \left(A {x}^{2} + B x + C\right)$

For the particular solution, we use trial solution:

${y}_{p} = {e}^{- x} \rho \left(x\right)$ where $\rho$ is a polynomial in $x$

$D {y}_{p} = - {e}^{- x} \rho + {e}^{- x} \rho ' = {e}^{- x} \left(- \rho + \rho '\right)$

${D}^{2} {y}_{p} ' = - {e}^{- x} \left(- \rho + \rho '\right) + {e}^{- x} \left(- \rho ' + \rho ' '\right) = {e}^{- x} \left(\rho - 2 \rho ' + \rho ' '\right)$

${D}^{3} {y}_{p} = - {e}^{- x} \left(\rho - 2 \rho ' + \rho ' '\right) + {e}^{- x} \left(\rho ' - 2 \rho ' ' + \rho ' ' '\right) = {e}^{- x} \left(- \rho + 3 \rho ' - 3 \rho ' ' + \rho ' ' '\right)$

And adding it all up means $\left({D}^{3} + 3 {D}^{2} + 3 D + 1\right) {y}_{p} = {x}^{2} {e}^{-} x$ amounts to ${e}^{- x} \rho ' ' ' = {x}^{2} {e}^{-} x$ so:

$\rho ' ' ' = {x}^{2}$

$\implies \rho ' ' = \frac{1}{3} {x}^{3} + \beta$

$\implies \rho ' = \frac{1}{12} {x}^{4} + \beta x + \gamma$

$\implies \rho = \frac{1}{50} {x}^{5} + \frac{\beta}{2} {x}^{2} + \gamma x + \delta$

The $\beta , \gamma , \delta$ terms are already in the complementary solution so we are left with:

$y = {y}_{c} + {y}_{p} = {e}^{- x} \left({x}^{5} / 60 + A {x}^{2} + B x + C\right)$