91Ó°ÊÓ

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$a=bâ‰$ $c$, therefore, the solution would be in the form of eq.(6.18). Now, let

$u=(D−2)y$

Therefore, the differential equation becomes

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

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

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

Solve further

$\begin{array}{l}=−8e^{−2}+{\mathrm{Î\pm }}_{1}u\\ =−8+{\mathrm{Î\pm }}_1{e}^{2x}\end{array}$

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

$y^{\text{'}}−2y=−8+{\mathrm{Î\pm }}_1{e}^{2x}$

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

$\begin{array}{c}I=∫−2dx\\ =−2x\\ e^I=e^{−2x}y{e}^{−2x}\\ =∫(−8+{\mathrm{Î\pm }}_1{e}^{2x})e^{−2x}dx\end{array}$

Solve further

$\begin{array}{l}=4e^{−2x}+{\mathrm{Î\pm }}_{1}x+{\mathrm{Î\pm }}_{2}y\\ =4+({\mathrm{Î\pm }}_{1}x+{\mathrm{Î\pm }}_2)e^{2x}\end{array}$

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.