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)].

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{16}\mathit{y}\mathbf{=}\mathbf{40}{\mathit{e}}^{\mathbf{4}\mathbf{x}}$

Short answer

The general solution of the differential equation is.$y=A{e}^{-4x}+B{e}^{4x}+5x{e}^{4x}$

Step-by-step solution

Given information

A differential equation is given as.$D(D+5)y=0$

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,

$\begin{array}{c}(D^2-16)y=40e^{4x}\\ (D+4)(D-4)y=40e^{4x}\end{array}$

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

$(D+4)(D-4)y=0$

The solution for this differential equation is in the form of eq.(5.11) because the roots of the auxiliary equation are not equal. That is,$y_c=A{e}^{-4x}+B{e}^{4x}$

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

$u=(D-4)y$

therefore, the differential equation becomes

$\begin{array}{c}(D+4)u=40e^{4x}\\ u^{\text{'}}+4u=40e^{4x}\end{array}$

General solution differential equation 

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

$\begin{array}{c}I=\int 4dx\\ I=4x\\ e^I=e^{4x}u{e}^I\\ =\int \left(40e^{4x}\right)e^{4x}dx\end{array}$

Solve further the equation

$\begin{array}{l}=5e^{8x}u\\ =5e^{4x}\end{array}$

Now, substitute this result in$u=(D-4)y$to get

$5e^{4x}=y^{\text{'}}-4y$

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

$\begin{array}{c}I=-\int 4dx\\ =-4x{y}_p{e}^I\\ =\int \left(5e^{4x}\right)e^{-4x}dx\\ =5x{y}_p\end{array}$

Further solve the equation

$=5x{e}^{4x}$

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

$y=A{e}^{-4x}+B{e}^{4x}+5x{e}^{4x}$

,,$\left[\text{DSolve}\right[\text{y},\text{x}\left[\text{x}\right]-16\text{y}\left[\text{x}\right]==40\text{E}\left(4\text{x}\right),\text{y}\left[\text{x}\right]$ get the same answer as above