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)].

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{6}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}\mathbf{12}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$

Short answer

The general solution of the differential equation is$y=(Ax+B)e^{-3x}+3e^{-x}$

Step-by-step solution

Given information

A differential equation is given as.$(D^2+16)y=0$

Step 2:Auxiliary equation

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

Roots of the auxiliary equation  

First, write the auxiliary equation

$\begin{array}{r}(D^2+6D+9)y=12e^{-x}\\ (D+3)(D+3)y=12e^{-x}\end{array}$

The complementary solution is corresponding to the same differential equation with zero right-hand side, that is$(D+3)(D+3)y=0$

and the solution for this differential equation is in the form of eq. (5.15) because the roots are equal. That is,

$y_c=(Ax+B)e^{-3x}$

Now, the particular- solution 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

$u=(D+3)y$

order of differential equation

Therefore, the differential equation becomes

$\begin{array}{l}(D+3)u=12e^{-x}\\ u^{\text{'}}+3u=\end{array}$

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

$\begin{array}{c}I=\int 3dx\\ I=3x\\ e^I=e^{3x}\end{array}$,

Solve further

$\begin{array}{c}u{e}^I=\int \left(12e^{-x}\right)e^{3x}dx\\ =6e^{2x}\\ u=6e^{-x}\end{array}$

General equation of differential equation

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

$6e^{-x}=y^{\text{'}}+3y$

which has become (again) a first order differential equation. find the solution of such an equation as follow

$\begin{array}{c}I=\int 3dx\\ I=3x\\ y_p{e}^I=\int \left(6e^{-x}\right)e^{3x}dx\end{array}$

Solve further the integral

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

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

$y=(Ax+B)e^{-3x}+3e^{-x}$

Write D Solve,$\left[y^{\text{'}},\left[x\right]+6y^{\text{'}}\left[x\right]+9y\left[x\right]==12E^{\wedge }(-x),y\left[x\right],x\right]$ get the same answer as above.