Second Order Linear Differential Equation $(t+3)^2y'' + (6t+18)y'+6y = e^t$ for $t > -3$

Question

Question States:

"Find the general solution to the following equation $$(t+3)^2y'' + (6t+18)y'+6y = e^t$$ for $t > -3$.

I am not sure how to proceed. First we can try to find 2 linearly independent solutions to the homogeneous equation, and use variation of parameters to find a particular solution and add all three up. How would I be able to solve the 2 equations for the homogeneous equation? I tried setting $y$ equal to a second degree polynomial but failed; similarly for setting $y$ equal to $e^{\lambda t}$.

Answer 1

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According to Maple, the general solution to the homogeneous DE is

$$ y \left( t \right) ={\frac {c_{{1}}{{\rm e}^{ 1/\left( t+3 \right) }} \left( 2\,t+7 \right) }{ \left( t+3 \right) ^{4}}}+{\frac {c_{{2}} \left( \left( -2\,t-7 \right) {{\rm e}^{ 1/\left( t+3 \right)}}{ \it Ei} \left( 1, \left( t+3 \right) ^{-1} \right) +{t}^{2}+7\,t+12 \right) }{ \left( t+3 \right) ^{4}}} $$

where $Ei$ is an exponential integral function. In particular, it is not elementary. You can then get a particular solution using variation of parameters, but it won't be pretty.

Answer 2

Using $x=\ln(t+3)$ for $t+3>0$ we find $(t+3)y'=\dfrac{dy}{dx}=Y'$ and $(t+3)^2y''=\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=Y''-Y'$ after simplification you get $$Y''+5Y'+6Y=0$$ that concludes general solution $Y=C_1e^{-2x}+C_2e^{-3x}$ or $$y=\frac{C_1}{(t+3)^2}+\frac{C_2}{(t+3)^3}$$

Answer 3

The equation can be written as $$(t+3)^2y''+6(t+3)y'+6y=e^t$$ Let's suppose that $y=(t+3)^k$ is a solution of the homogeneous DE $$(t+3)^2y''+6(t+3)y'+6y=0\tag{1}$$ i.e. \begin{align*} &&k(k-1)(t+3)^k+6k(t+3)^k+6(t+3)^k&=0\quad\text{for all }t>-3\\ \iff&&(k^2+5k+6)(t+3)^k&=0\quad\text{for all }t>-3\\ \iff&&k&\in\{-2,-3\} \end{align*}

Then, $y_c=c_1(t+3)^{-2}+c_2(t+3)^{-3},\;\;t>-3\;$ is a solution of the homogeneous $(1)$, where $c_1$ and $c_2$ are arbitrary constants. Now proceed for variation of parameters: look for two functions $u_1=u_1(t)$ and $u_2=u_2(t)$ such that $$(t+3)^{-2}u_1'+(t+3)^{-3}u_2'=0\qquad\text{and}\qquad -2(t+3)^{-3}u_1'-3(t+3)^{-4}u_2'=e^t(t+3)^{-2}$$ Hence $$u_1'=\frac{\begin{vmatrix}0&(t+3)^{-3}\\e^t(t+3)^{-2}&-3(t+3)^{-4}\end{vmatrix}}{\begin{vmatrix}(t+3)^{-2}&(t+3)^{-3}\\-2(t+3)^{-3}&-3(t+3)^{-4}\end{vmatrix}}=\frac{-e^t(t+3)^{-5}}{-(t+3)^{-6}}=(t+3)e^t\quad\implies\quad u_1=(t+2)e^t$$ $$u_2'=\frac{\begin{vmatrix}(t+3)^{-2}&0\\-2(t+3)^{-3}&e^t(t+3)^{-2}\end{vmatrix}}{\begin{vmatrix}(t+3)^{-2}&(t+3)^{-3}\\-2(t+3)^{-3}&-3(t+3)^{-4}\end{vmatrix}}=\frac{e^t(t+3)^{-4}}{-(t+3)^{-6}}=-(t+3)^2e^t\implies u_2=-(t^2+4t+5)e^t$$ Then, the solution is $$y=c_1(t+3)^{-2}+c_2(t+3)^{-3}+(t+2)(t+3)^{-2}e^t-(t^2+4t+5)(t+3)^{-3}e^t$$ which simplifies to $$\boxed{y=c_1(t+3)^{-2}+c_2(t+3)^{-3}+(t+1)(t+3)^{-3}e^t}$$