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)].

$y^{\text{'}\text{'}}-4y=10$

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 $a\ne$$b\ne 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

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

Solve further

.$y^{\text{'}}+2y=-5+{\alpha }_1{e}^{2x}$

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

Solve

$\begin{array}{c}I=\int 2dx\\ I=2x\\ e^I=e^{2x}\\ y{e}^{2x}=\int (-5+{\alpha }_1{e}^{2x})e^{2x}dx\end{array}$

Solve further

$\begin{array}{l}=-\frac{5}{2}{e}^{2x}+{\alpha }_2{e}^{4x}+{\alpha }_{3}y\\ =-\frac{5}{2}+{\alpha }_2{e}^{2x}+{\alpha }_3{e}^{-2x}\end{array}$

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