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

a) By inspection, find a particular solution of
$$\displaystyle{y}{''}+{2}{y}={10}$$
b) By inspection, find a particular solution of
$$\displaystyle{y}{''}+{2}{y}=-{4}{x}$$
c) Find a particular solution of
$$\displaystyle{y}{''}+{2}{y}=-{4}{x}+{10}$$
d) Find a particular solution of
$$\displaystyle{y}{''}+{2}{y}={8}{x}+{5}$$

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Step 1
a) By inspection we see that the particular solution of the DE
$$\displaystyle{y}{''}+{2}{y}={10}$$ is $$\displaystyle{y}_{{{p}{1}}}={5}$$
$$\displaystyle\because{y}_{{{p}{1}}}{''}+{2}{y}_{{{p}{1}}}={0}+{2}{\left({5}\right)}={10}$$
b) By inspection we see that the particular solution of the DE
$$\displaystyle{y}{''}+{2}{y}=-{4}{x}$$ is $$\displaystyle{y}_{{{p}{2}}}=-{2}{x}$$
$$\displaystyle\because{y}_{{{p}{2}}}{''}+{2}{y}_{{{p}{2}}}={0}+{2}{\left(-{2}{x}\right)}=-{4}{x}$$
c) By Superposition principle we see that the particular solution of the DE
$$\displaystyle{y}{''}+{2}{y}=-{4}{x}+{10}$$ is the superposition of the solutions in parts (a) and (b)
$$\displaystyle\therefore{y}_{{{p}}}={y}_{{{p}{1}}}+{y}_{{{p}{2}}}\rightarrow{y}_{{{p}}}={5}-{2}{x}$$
$$\displaystyle\because{y}_{{{p}}}{''}+{2}{y}_{{{p}}}={0}+{2}{\left({5}-{2}{x}\right)}=-{4}{x}+{10}$$
d) Be Superposition principle we see that the particular solution of the DE $$\displaystyle{y}{''}+{2}{y}={8}{x}+{5}=-{2}{\left(-{4}{x}\right)}+{\frac{{{1}}}{{{2}}}}{\left({10}\right)}$$ is the superposition of the solutions in parts (a) and (b)
$$\displaystyle\therefore{y}_{{{p}}}={\frac{{{1}}}{{{2}}}}{y}_{{{p}{1}}}-{2}{y}_{{{p}{2}}}\rightarrow{y}_{{{p}}}={\frac{{{1}}}{{{2}}}}{\left({5}\right)}-{2}{\left(-{2}{x}\right)}$$
$$\displaystyle\rightarrow{y}_{{{p}}}={\frac{{{5}}}{{{2}}}}+{4}{x}$$
$$\displaystyle\because{y}_{{{p}}}{''}+{2}{y}_{{{p}}}={0}+{2}{\left({\frac{{{5}}}{{{2}}}}+{4}{x}\right)}={8}{x}+{5}$$
a) $$\displaystyle{y}_{{{p}{1}}}={5}$$
b) $$\displaystyle{y}_{{{p}{2}}}=-{2}{x}$$
c) $$\displaystyle{y}_{{{p}}}={5}-{2}{x}$$
d) $$\displaystyle{y}_{{{p}}}={\frac{{{5}}}{{{2}}}}+{4}{x}$$
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Ched1950

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 $$\displaystyle{y}'={y}{''}={0}$$. Substituting these in the equation gives.
$$\displaystyle{y}{''}+{2}{y}={10}$$
$$\displaystyle{0}+{2}{c}={10}$$
$$\displaystyle{c}={5}$$
Hence, by inspection, $$\displaystyle{y}={5}$$ is the particular solution of $$\displaystyle{y}{''}+{2}{y}={10}$$
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 $$\displaystyle{y}={c}{x}$$ as the particular solution of the given equation. So, we have $$\displaystyle{y}'={c},\ {y}{''}={0}.$$
Substituting these in the equation gives
$$\displaystyle{y}{''}+{2}{y}=-{4}$$
$$\displaystyle{0}+{2}{c}{x}=-{4}{x}$$
$$\displaystyle{c}=-{2}$$
Hence, by inspection, $$\displaystyle{y}=-{2}{x}$$ is the particular solution of $$\displaystyle{y}{''}+{2}{y}=-{4}{x}$$
Step 2
c) From part (a), we saw that $$\displaystyle{y}_{{{p}{1}}}={5}$$ is the particular solution of
$$\displaystyle{y}{''}+{2}{y}={10}$$
and from part (b), $$\displaystyle{y}_{{{p}{2}}}=-{2}{x}$$ is the perticular solution of
$$\displaystyle{y}{''}+{2}{y}=-{4}{x}$$
Therefore, it follows from (13) of Theorem 4.1.7 that the solution of
$$\displaystyle{y}{''}+{2}{y}=\underbrace{{-{4}{x}}}_{{{{g}_{{{2}}}{\left({x}\right)}}}}+\underbrace{{{10}}}_{{{{g}_{{{1}}}{\left({x}\right)}}}}$$
is the superposition of $$\displaystyle{y}_{{{p}{2}}}$$ and $$\displaystyle{y}_{{{p}{1}}}$$, i.e.
$$\displaystyle{y}={y}_{{{p}{2}}}+{y}_{{{p}{1}}}$$
$$\displaystyle=-{2}{x}+{5}$$ Step 3
d) Similarly, the given equation can be written as
$$\displaystyle{y}{''}+{2}{y}={8}{x}+{5}$$
$$\displaystyle=-{2}{\left(-{4}{x}\right)}+{\left({\frac{{{1}}}{{{2}}}}\right)}{\left({10}\right)}$$
$$\displaystyle=-{2}\underbrace{{{\left(-{4}{x}\right)}}}_{{{{g}_{{{2}}}{\left({x}\right)}}}}+{\left({\frac{{{1}}}{{{2}}}}\right)}\underbrace{{{\left({10}\right)}}}_{{{{g}_{{{1}}}{\left({x}\right)}}}}$$
From part (a), we saw that $$\displaystyle{y}_{{{p}{1}}}={5}$$ is the particular solution of
$$\displaystyle{y}{''}+{2}{y}={10}$$
and from part (b) $$\displaystyle{y}_{{{p}{2}}}=-{2}{x}$$ is the particular solution of
$$\displaystyle{y}{''}+{2}{y}=-{4}{x}$$
Therefore, it follows from (13) of Theorem 4.1.7 that the solution of
$$\displaystyle{y}{''}+{2}{y}=-{2}\underbrace{{{\left(-{4}{x}\right)}}}_{{{{g}_{{{2}}}{\left({x}\right)}}}}+{\left({\frac{{{1}}}{{{2}}}}\right)}\underbrace{{{\left({10}\right)}}}_{{{{g}_{{{1}}}{\left({x}\right)}}}}$$
is the superposition of $$-2y_{p2}$$ and $$\displaystyle{\frac{{{1}}}{{{2}}}}{y}_{{{p}{1}}}$$
$$\displaystyle{y}=-{2}{y}_{{{p}{2}}}+{\frac{{{1}}}{{{2}}}}{y}_{{{p}{1}}}$$
$$\displaystyle=-{2}{\left(-{2}{x}\right)}+{\left({\frac{{{1}}}{{{2}}}}\right)}{\left({5}\right)}$$
$$\displaystyle=-{2}{\left(-{2}{x}\right)}+{\left({\frac{{{1}}}{{{2}}}}\right)}{\left({5}\right)}$$
$$\displaystyle={4}{x}+{\frac{{{5}}}{{{2}}}}$$