Question

Use the method of variation of parameters to determine the general solution of the given differential equation.

y′′′−y′=5t

Use C1, C2, C3, ... for the constants of integration.

y(t) = _______

Answer 1

The given differential equation can be written as $(D^{3}-D)y = 5t,~ \text{where}~ D\equiv \dfrac{d}{dt}.$

The auxiliary equation is, $m^{3}-m=0.$ Solving, we get $m = 0,1,-1.$

The complementary function is, $y_{c} = c_{1}+c_{2}e^{t}+c_{3}e^{-t} = c_{1}f_{1}+c_{2}f_{2}+c_{3}f_{3}$

The particular integral is $y_{p}=Pf_{1}+Qf_{2}+Rf_{3},$ where

$P=\displaystyle\int \frac{W_{1}}{W(f_{1},f_{2},f_{3})}dx; Q=\displaystyle\int \frac{W_{2}}{W(f_{1},f_{2},f_{3})}dx;\\~\\ R=\displaystyle\int \frac{W_{3}}{W(f_{1},f_{2},f_{3})}dx;$

and $W_{i}$ is the determinant obtained from the Wronskian $W(f_{1},f_{2},f_{3})$ by replacing the $i^{th}$ - column by the vector $\begin{pmatrix} 0 \\ 0 \\ g(t) \end{pmatrix}$ where $g(t) = 5t.$

Now,

$W(f_{1},f_{2},f_{3}) = \begin{vmatrix} 1 & e^{t} & e^{- t} \\ 0 & e^{t} &- e^{- t} \\0 & e^{t} & e^{- t} \end{vmatrix} = 2\\~\\ W_{1} = \begin{vmatrix} 0 & e^{t} & e^{- t} \\ 0 & e^{t} &- e^{- t} \\5t & e^{t} & e^{- t} \end{vmatrix} = -10t\\~\\ W_{2} = \begin{vmatrix} 1 & 0 & e^{- t} \\ 0 & 0 &- e^{- t} \\0 & 5t & e^{- t} \end{vmatrix} = 5te^{-t}\\~\\ W_{3} = \begin{vmatrix} 1 & e^{t} & 0 \\ 0 & e^{t} & 0 \\0 & e^{t} &5t \end{vmatrix} = 5te^{t}$

Therefore,

$P = \displaystyle\int -5t~ dt = -\frac{5t^{2}}{2}\\~\\ Q= \displaystyle \frac{5}{2}\int te^{-t}~ dt = \frac{5}{2}(-te^{-t}-e^{-t}) = -\frac{5}{2}e^{-t}(t+1)\\~\\ R= \displaystyle \frac{5}{2}\int te^{t}~ dt = \frac{5}{2}(te^{t}-e^{t}) = \frac{5}{2}e^{t}(t-1)$

Hence,

$\begin{aligned} y_{p} &=Pf_{1}+Qf_{2}+Rf_{3}\\ &=-\frac{5t^{2}}{2}-\frac{5}{2}e^{-t}(t+1) \cdot e^{t} + \frac{5}{2}e^{t}(t-1) \cdot e^{-t}\\ &=-\frac{5t^{2}}{2}-\frac{5}{2}(t+1) + \frac{5}{2}(t-1)\\ &=-\frac{5t^{2}}{2} - 5 \end{aligned}$

The general solution is

$\begin{aligned} y = y_{c}+y_{p} &= c_{1}+c_{2}e^{t}+c_{3}e^{-t}-\dfrac{5t^{2}}{2}-5\\ y&= C_{1}+c_{2}e^{t}+c_{3}e^{-t}-\dfrac{5t^{2}}{2}, \text{where}~C_{1}=c_{1}-5. \end{aligned}$