a) By inspection, find a particular solution of y''+2y=10 b) By inspection,

Step 1
The simplest method for finding a particular solution is to try a solution of the same form as the right hand side of the nonhomogeneous equation, i.e. if it's a polynomial of degree n, we try a general polynomial of degree n and see if we can match the coefficients.
Since, the right hand side of the given equation is a constant, let us consider the constant function ${\displaystyle y=c}$ as the particular solution of the given equation. So, we have ${\displaystyle y^{\prime }=y{}^{″}=0}$. Substituting these in the equation gives.
${\displaystyle y{}^{″}+2y=10}$
${\displaystyle 0+2c=10}$
${\displaystyle c=5}$
Hence, by inspection, ${\displaystyle y=5}$ is the particular solution of ${\displaystyle y{}^{″}+2y=10}$
b) Now, for this exercise, since, the right hand side of the given equation is a polynomial of degree one, let us consider the function ${\displaystyle y=cx}$ as the particular solution of the given equation. So, we have ${\displaystyle y^{\prime }=c,y{}^{″}=0.}$
Substituting these in the equation gives
${\displaystyle y{}^{″}+2y=-4}$
${\displaystyle 0+2cx=-4x}$
${\displaystyle c=-2}$
Hence, by inspection, ${\displaystyle y=-2x}$ is the particular solution of ${\displaystyle y{}^{″}+2y=-4x}$
Step 2
c) From part (a), we saw that ${\displaystyle y_{p1}=5}$ is the particular solution of
${\displaystyle y{}^{″}+2y=10}$
and from part (b), ${\displaystyle y_{p2}=-2x}$ is the perticular solution of
${\displaystyle y{}^{″}+2y=-4x}$
Therefore, it follows from (13) of Theorem 4.1.7 that the solution of
${\displaystyle y{}^{″}+2y=\underset{g_2\left(x\right)}{\underbrace{-4x}}+\underset{g_1\left(x\right)}{\underbrace{10}}}$
is the superposition of ${\displaystyle y_{p2}}$ and ${\displaystyle y_{p1}}$, i.e.
${\displaystyle y=y_{p2}+y_{p1}}$
${\displaystyle =-2x+5}$ Step 3
d) Similarly, the given equation can be written as
${\displaystyle y{}^{″}+2y=8x+5}$
${\displaystyle =-2(-4x)+\left(\frac{1}{2}\right)\left(10\right)}$
${\displaystyle =-2\underset{g_2\left(x\right)}{\underbrace{(-4x)}}+\left(\frac{1}{2}\right)\underset{g_1\left(x\right)}{\underbrace{\left(10\right)}}}$
From part (a), we saw that ${\displaystyle y_{p1}=5}$ is the particular solution of
${\displaystyle y{}^{″}+2y=10}$
and from part (b) ${\displaystyle y_{p2}=-2x}$ is the particular solution of
${\displaystyle y{}^{″}+2y=-4x}$
Therefore, it follows from (13) of Theorem 4.1.7 that the solution of
${\displaystyle y{}^{″}+2y=-2\underset{g_2\left(x\right)}{\underbrace{(-4x)}}+\left(\frac{1}{2}\right)\underset{g_1\left(x\right)}{\underbrace{\left(10\right)}}}$
is the superposition of