91Ó°ÊÓ

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

-Auxiliary equation:

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

-General form of the auxiliary equation

$\mathbf{(}\mathit{D}\mathbf{−}\mathit{a}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{−}\mathit{b}\mathbf{)}\mathbf{=}\mathit{k}{\mathit{e}}^{\mathbf{c}\mathbf{x}}$

First, write the auxiliary equation

$(D+1)(D−3)=24e^{−3x}$

The complementary solution is corresponding to the same differential equation with zero right-hand side, that is

$(D+1)(D−3)y=0$

and the solution for this differential equation is in the form of eq.(5.11) because the roots are not equal.

Now, the particular- solution $y_p$could be found by successive integration of two first order equations (need to omit the integration constant each time to get the particular- solution). Let

$u=(D−3)y$

Therefore, the differential equation becomes

$\begin{array}{l}(D+1)u=24e^{−3x}\\ u^{\text{'}}+u=24e^{−3x}\end{array}$

This is a first order differential equation, and solve it by making use of eq., (3.4) and eq., (3.9) (remember, drop the integration constants), that is

$\begin{array}{c}I=∫dx\\ I=x\\ \text{ }{e}^I=e^{x}u\\ e^I=∫\left(24e^{−3x}\right)e^{x}dx\end{array}$

Solve further the intergal

$u=−12e^{−3x}$

Now, substitute this result in ,$u=(D−3)y$

$−12e^{−3x}=y^{\text{'}}−3y$Which is become (again) a first order differential equation.Find the solution of such an equation as follow

$\begin{array}{l}I=−∫3dx\\ I=−3x\\ e^I=e^{−}{y}_p\\ e^I=∫(−12e^{−3x})e^{−3x}dx\end{array}$

Solve further the integral

$\begin{array}{l}=2e^{−6x}{y}_p\\ =2e^{−3x}\end{array}$

Therefore, the general solution of the differential equation is,$y=y_c+y_p$ that is

$y={\mathrm{Î\pm }}_1{e}^{−x}+{\mathrm{Î\pm }}_2{e}^{3x}+2e^{−3x}$

,$\left[y^{\text{'}},\left[x\right]−2y^{\text{'}}\left[x\right]−3y\left[x\right]==24E^{âˆ\S }(−3x),y\left[x\right],x\right]$ get the same answer as above.