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

mronjo7n 2021-11-22 Answered
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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Expert Answer

oces3y
Answered 2021-11-23 Author has 57 answers
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)
\(\displaystyle\therefore{y}_{{{p}}}={\frac{{{1}}}{{{2}}}}{y}_{{{p}{1}}}-{2}{y}_{{{p}{2}}}\rightarrow{y}_{{{p}}}={\frac{{{1}}}{{{2}}}}{\left({5}\right)}-{2}{\left(-{2}{x}\right)}\)
\(\displaystyle\rightarrow{y}_{{{p}}}={\frac{{{5}}}{{{2}}}}+{4}{x}\)
\(\displaystyle\because{y}_{{{p}}}{''}+{2}{y}_{{{p}}}={0}+{2}{\left({\frac{{{5}}}{{{2}}}}+{4}{x}\right)}={8}{x}+{5}\)
Answer:
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
Answered 2021-11-24 Author has 1450 answers

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 \(\displaystyle{\frac{{{1}}}{{{2}}}}{y}_{{{p}{1}}}\)
\(\displaystyle{y}=-{2}{y}_{{{p}{2}}}+{\frac{{{1}}}{{{2}}}}{y}_{{{p}{1}}}\)
\(\displaystyle=-{2}{\left(-{2}{x}\right)}+{\left({\frac{{{1}}}{{{2}}}}\right)}{\left({5}\right)}\)
\(\displaystyle=-{2}{\left(-{2}{x}\right)}+{\left({\frac{{{1}}}{{{2}}}}\right)}{\left({5}\right)}\)
\(\displaystyle={4}{x}+{\frac{{{5}}}{{{2}}}}\)

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