Solution to the linear differential equation using successive integration of two first-order equations Reading through Boas' mathematical methods books, one example went as follows:
$$y^{\prime\prime}+y^{\prime}-2y=e^x\tag1$$
$$\left(D^2+D-2\right)y=e^x$$
$$\left({D-1}\right)\left({D+2}\right)y=e^x\tag2$$
let $u=\left({D+2}\right)y=e^x$. Then the differential equation becomes
$$\left({D-1}\right)u=e^x$$
$$u^{\prime\prime}-u^{\prime}=e^x\tag3$$
The solution of Eq. (3) would then be
$$ue^{-x}=\int{e^{-x}e^{x}}dx=x+c_1$$
$$u=xe^{x}+c_1e^{x}\tag4$$
Then the differential equation for y becomes
$$\left({D+2}\right)y=xe^{x}+c_1e^{x}$$
$$y^{\prime}+2y=xe^{x}+c_1e^{x}$$
This is again a linear first-order equation which can be solved as follows:
$$ye^{2x}=\int{e^{2x}\left(xe^{x}+c_1e^{x}\right)}dx=\frac13xe^{3x}-\frac19e^{3x}+\frac13{c_1}e^{3x}+c_2\tag{5.1}$$
$$ye^{2x}=\frac13xe^{3x}+{c_1}^{\prime}e^{3x}+c_2\tag{5.2}$$
$$y=\frac13xe^{x}+{c_1}^{\prime}e^{x}+{c_2}e^{-2x}\tag6$$
From Eq. (5.1) to Eq. (5.2), it could be stated that
$$\frac13{c_1}e^{3x}-\frac19e^{3x}={c_1}^{\prime}e^{3x}$$
$$\frac13{c_1}-\frac19={c_1}^{\prime}$$
But I am not quite sure as to how that's done. Any help would be much appreciated. Thank you!

$$ye^{2x}=\int{e^{2x}\left(xe^{x}+c_1e^{x}\right)}dx$$
$$\implies ye^{2x}=\frac13xe^{3x}-\frac19e^{3x}+\frac13{c_1}e^{3x}+c_2\tag{5.1}$$
Let this equality $(5.1)$ can be written as
$$ye^{2x}=\frac13xe^{3x}+{c_1}^{\prime}e^{3x}+c_2\tag{5.2}$$
Comparing equation $(5.1)$ and $(5.2)$ ,$~\color{red}{\text{(that means comparing the coefficients of similar powers of $e$ in equation $(5.1)$ and $(5.2)$)}}~$ we have
$$c_1~'=\frac13{c_1}-\frac19$$
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Here the differential equation is $$y^{\prime\prime}+y^{\prime}-2y=e^x\tag1$$
$$\implies \left(D^2+D-2\right)y=e^x\qquad \text{where}\quad D\equiv \frac{d}{dx}$$
Taking the equation $$y^{\prime\prime}+y^{\prime}-2y=0\tag2$$ and putting $~y=e^{mx}~$ in it, then we have
$$m^2+m-2=0$$called auxiliary equation. Roots of the auxiliary equation is $~m=-2,~1~$.
Then the solution of $(2)$ is $$y~=~c_1~ e^x ~+~c_2~e^{-2x}\qquad \text{where}\quad c_1,~c_2~\text{are constants.}$$
This $$c_1~ e^x ~+~c_2~e^{-2x}$$  is called the complementary function (C.F.) of $(1)$.
For non-homogeneous part, we have to find the particular integral (P.I.)
P.I.$~=\frac{1}{D^2+D-2}~e^x=\frac{1}{3}~\left(\frac{1}{D-1}-\frac{1}{D+2}\right)~e^x=\frac{1}{3}~\left(x~e^x-\frac{1}{3}~e^x\right)$
Hence the general solution of $(1)$ is $$y(x)~=~C.F.~+~P.I.$$
$$\implies y(x)=~c_1~ e^x ~+~c_2~e^{-2x}~+~\frac{1}{3}~\left(x~e^x-\frac{1}{3}~e^x\right)$$
$$\implies y(x)=\frac13xe^{x}+{c_1}^{\prime}e^{x}+{c_2}e^{-2x}$$where $~{c_1}^{\prime}={c_1}-\frac19~~$ and $~ c_2~$ are constants.


Consider a differential equation of the form $f(D)y=X$
If $X=e^{ax}$, then
$1.$ P.I.$\quad = \frac{1}{f(D)}e^{ax}=\frac{e^{ax}}{f(a)}$, if $f(a)\neq 0$
$2.$ P.I.$\quad =\frac{1}{(D-a)^n}e^{ax}=\frac{x^n}{n!}e^{ax}$

