91Ó°ÊÓ

A differential equation is given as$y^{\text{'}\text{'}}+y^{\text{'}}−2y=0$

-Auxiliary equation:

Auxiliary equation is an algebraicequation of degreeupon which depends thesolution of a given nth-order differential equation or difference equation.

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

write the auxiliary equation

$\begin{array}{r}(D^2−4)y=10\\ (D−2)(D+2)y=10\end{array}$

Ifcompare it to the general form of the auxiliary equation$(D−a)(D−b)=k{e}^{cx}$here, $a=2,b=−2$and,$c=0$so, it is clear that $aâ‰$$b≠c$, therefore, the solution would be in the form of eq.(6.18). Now, let

$u=(D+2)y$

therefore, the differential equation becomes

$\begin{array}{l}(D−2)u=10\\ u^{\text{'}}−2u=10\end{array}$

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

$\begin{array}{c}I=∫−2dx\\ =−2x{e}^I\\ =u{e}^{−2x}\\ =∫\left(10\right)e^{−2x}dx\end{array}$

Solve further

.$y^{\text{'}}+2y=−5+{\mathrm{Î\pm }}_1{e}^{2x}$

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

Solve

$\begin{array}{c}I=∫2dx\\ I=2x\\ e^I=e^{2x}\\ y{e}^{2x}=∫(−5+{\mathrm{Î\pm }}_1{e}^{2x})e^{2x}dx\end{array}$

Solve further

$\begin{array}{l}=−\frac{5}{2}{e}^{2x}+{\mathrm{Î\pm }}_2{e}^{4x}+{\mathrm{Î\pm }}_{3}y\\ =−\frac{5}{2}+{\mathrm{Î\pm }}_2{e}^{2x}+{\mathrm{Î\pm }}_3{e}^{−2x}\end{array}$

By the computer software $[y,[x]−4y[x]==10,y[x],x]$, and get the same answer as above.