Question

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)].

$\mathbf{(}\mathit{D}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathbf{3}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{24}{\mathit{e}}^{\mathbf{-}\mathbf{3}\mathbf{x}}$

Short answer

The general solution of the differential equationis.$y={\alpha }_1{e}^{-x}+{\alpha }_2{e}^{3x}+2e^{-3x}$

Step-by-step solution

Given information

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

Auxiliary equation

-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

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

Solve for differential equation

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.

Order of differential equation

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

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

Solve further the intergal

$u=-12e^{-3x}$

General solution of the differential equation  

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

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

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

$y={\alpha }_1{e}^{-x}+{\alpha }_2{e}^{3x}+2e^{-3x}$

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