Find the general solution of the following differential equations (complementary function particular solution). Find the solution by inspection or by (6.18), (6.23), or (6.24). Also find a computer solution and reconcile differences if necessary, noticing especially whether the solution is in simplest form [see (6.26) and the discussion after (6.15)].

${(D-2)}^{2}y=16$

A differential equation is given as .$y^{\text{'}\text{'}}-4y^{\text{'}}+4y=0$

-Auxiliary equation:

Auxiliary equation is an algebraic equation of degreeupon which depends the solution of a given nth-order differential equation or difference equation.

-General form of the auxiliary equation$(D-a)(D-b)=k{e}^{cx}$

First,write the auxiliary equation

$(D-2)(D-2)y=16$

Compare it to the general form of the auxiliary equation $(D-a)(D-b)=k{e}^{cx}$where , $a=2,b=2$and $c=0$, so, it is clear that$a=b\ne$ $c$, therefore, the solution would be in the form of eq.(6.18). Now, let

$u=(D-2)y$

Therefore, the differential equation becomes

$\begin{array}{c}(D-2)u=16\\ u^{\text{'}}-2u=16\end{array}$

Which has become first order linear differential equation, and solve it using eq.(3.4) and eq.(3.9), that is

$\begin{array}{c}I=\int -2dx\\ =-2x\\ e^I=u{e}^{-2x}\\ =\int \left(16\right)e^{-2x}dx\end{array}$

Solve further

$\begin{array}{l}=-8e^{-2}+{\alpha }_{1}u\\ =-8+{\alpha }_1{e}^{2x}\end{array}$

Now, substitute$u=-8+{\alpha }_1{e}^{2x}$ in the differential equation, $(D-2)y=u$get,

$y^{\text{'}}-2y=-8+{\alpha }_1{e}^{2x}$

Which is again first order differential equation and solve it in the same way.

$\begin{array}{c}I=\int -2dx\\ =-2x\\ e^I=e^{-2x}y{e}^{-2x}\\ =\int (-8+{\alpha }_1{e}^{2x})e^{-2x}dx\end{array}$

Solve further

$\begin{array}{l}=4e^{-2x}+{\alpha }_{1}x+{\alpha }_{2}y\\ =4+({\alpha }_{1}x+{\alpha }_2)e^{2x}\end{array}$

By the computer software ,$\left[y^{\text{'}},\left[x\right]-4y^{\text{'}}\left[x\right]+4y\left[x\right]==16,y\left[x\right],x\right]$ get the same answer as above.