I'm trying to find particular solution of the second-order linear

protisvitfc 2021-11-19 Answered
I'm trying to find particular solution of the second-order linear equation but I can't find \(\displaystyle{y}_{{1}}\) and \(\displaystyle{y}_{{2}}\) according to \(\displaystyle{y}={c}_{{1}}{y}_{{1}}+{c}_{{2}}{y}_{{2}}\)
\(\displaystyle{x}^{{2}}{y}{''}-{2}{x}{y}'+{2}{y}={0},\ {y}{\left({1}\right)}={3},\ {y}'{\left({1}\right)}={1}\)
If r is used, \(\displaystyle{x}^{{2}}{r}^{{2}}-{2}{x}{r}=-{2}\) then \(\displaystyle{x}{r}{\left({x}{r}-{2}\right)}=-{2}\), I can't go on from there to find \(\displaystyle{y}_{{1}}\) and \(\displaystyle{y}_{{2}}\)

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Expert Answer

Roger Noah
Answered 2021-11-20 Author has 1072 answers
Assume \(\displaystyle{y}={x}^{{r}}\) then if \(\displaystyle{y}={x}^{{r}}\) then
\(\displaystyle{y}'={r}{x}^{{{r}-{1}}}\)
And
\(\displaystyle{y}{''}={r}{\left({r}-{1}\right)}{x}^{{{r}-{2}}}\)
Substitute in
\(\displaystyle{x}^{{2}}{y}{''}-{2}{x}{y}'+{2}{y}={0}\)
to get
\(\displaystyle{r}{\left({r}-{1}\right)}{x}^{{r}}-{2}{r}{x}^{{r}}+{2}{x}^{{r}}={0}\)
divide by y to get
\(\displaystyle{r}{\left({r}-{1}\right)}-{2}{r}+{2}={\left({r}-{1}\right)}{\left({r}-{2}\right)}={0}\)
therefore
\(\displaystyle{y}={a}{x}+{b}{x}^{{2}}\)
Solve
\(\displaystyle{y}{\left({1}\right)}={a}+{b}={3}\)
\(\displaystyle{y}'{\left({1}\right)}={a}+{2}{b}={1}\)
to get \(\displaystyle{y}={5}{x}-{2}{x}^{{2}}\)
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Charles Wee
Answered 2021-11-21 Author has 454 answers

\(\displaystyle{y}{''}{\left({x}\right)}{x}^{{2}}+{2}{y}'{\left({x}\right)}{x}+{2}{y}{\left({x}\right)}={0}\)
Assume a solution to this Euler-Cauchy equation will be proportional to \(\displaystyle{e}^{\lambda}\) for some constant \(\displaystyle\lambda\)
Substitute \(\displaystyle{y}={x}^{{\lambda}}\) into the differential equation:
\(\displaystyle{x}^{{2}}{\frac{{{d}^{{2}}}}{{{\left.{d}{x}\right.}^{{2}}}}}{\left({x}^{\lambda}\right)}+{2}{x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}^{\lambda}\right)}+{2}{x}^{\lambda}={0}\)
Substitute \(\displaystyle{\frac{{{d}^{{2}}}}{{{\left.{d}{x}\right.}^{{2}}}}}{\left({x}^{\lambda}\right)}={\left(\lambda-{1}\right)}{x}^{{\lambda-{2}}}\) and \(\frac{d}{dx}(x^\lambda)=\lambda x^{\lambda-1}\)
\(\displaystyle\lambda^{{2}}{x}^{\lambda}+\lambda{x}^{\lambda}+{2}{x}^{\lambda}={0}\)
\(\displaystyle{x}^{\lambda}{\left(\lambda^{{2}}+\lambda+{2}\right)}={0}\)
Assuming \(\displaystyle{x}\ne{0}\), the zeros must come from the polynomial:
\(\displaystyle\lambda^{{2}}+\lambda+{2}={0}\)
\(\displaystyle\lambda=-{\frac{{{1}}}{{{2}}}}\pm{\frac{{{i}\sqrt{{{7}}}}}{{{2}}}}\)
The roots \(\displaystyle\lambda=-{\frac{{{1}}}{{{2}}}}\pm{\frac{{{i}\sqrt{{{2}}}}}{{{2}}}}\) give \(\displaystyle{y}_{{1}}{\left({x}\right)}={C}_{{1}}{x}^{{-{\frac{{{1}}}{{{2}}}}+{\frac{{{i}\sqrt{{{7}}}}}{{{2}}}}}}+{C}_{{2}}{x}^{{-{\frac{{{1}}}{{{2}}}}-{\frac{{{i}\sqrt{{{7}}}}}{{{2}}}}}}\) as solutions, where \(\displaystyle{C}_{{1}}\) and \(\displaystyle{C}_{{2}}\) are arbitrary constants. The general solution is the sum of the above solutions:
\(\displaystyle{y}{\left({x}\right)}={y}_{{1}}{\left({x}\right)}+{y}_{{2}}{\left({x}\right)}={C}_{{1}}{x}^{{-{\frac{{{1}}}{{{2}}}}+{\frac{{{i}\sqrt{{{7}}}}}{{{2}}}}+{C}_{{2}}{x}^{{-{\frac{{{1}}}{{{2}}}}-{\frac{{{i}\sqrt{{{7}}}}}{{{2}}}}}}}}\)
Using \(\displaystyle{x}^{\lambda}={e}^{{\lambda{\ln{{\left({x}\right)}}}}}\), apply Euler's identity \(\displaystyle{e}^{{\alpha+{b}{i}}}={e}^{\alpha}{\cos{{\left({b}\right)}}}+{i}{e}^{\alpha}{\sin{{\left({b}\right)}}}\)
\(\displaystyle{y}{\left({x}\right)}={\frac{{{\left({C}_{{1}}+{C}_{{2}}\right)}{\cos{{\left({\frac{{{1}}}{{{2}}}}\sqrt{{{7}}}{\ln{{\left({x}\right)}}}\right)}}}}}{{\sqrt{{{x}}}}}}+{\frac{{{i}{\left({C}_{{1}}-{C}_{{2}}\right)}{\sin{{\left({\frac{{{1}}}{{{2}}}}\sqrt{{{7}}}{\ln{{\left({x}\right)}}}\right)}}}}}{{\sqrt{{{x}}}}}}\)
\(\displaystyle{y}{\left({x}\right)}={\frac{{{C}_{{1}}{\cos{{\left({\frac{{{1}}}{{{2}}}}\sqrt{{{7}}}{\ln{{\left({x}\right)}}}\right)}}}}}{{\sqrt{{{x}}}}}}+{\frac{{{C}_{{2}}{\sin{{\left({\frac{{{1}}}{{{2}}}}\sqrt{{{7}}}{\ln{{\left({x}\right)}}}\right)}}}}}{{\sqrt{{{x}}}}}}\)
Now using some algebra to find \(\displaystyle{C}_{{1}}\) and \(\displaystyle{C}_{{2}}\) with \(\displaystyle{y}{\left({1}\right)}={3}\) and \(\displaystyle{y}'{\left({1}\right)}={1}\):
\(\displaystyle{y}{\left({x}\right)}={\frac{{{21}{\cos{{\left({\frac{{{1}}}{{{2}}}}\sqrt{{{7}}}{\ln{{\left({x}\right)}}}\right)}}}+{5}\sqrt{{{7}}}{\sin{{\left({\frac{{{1}}}{{{2}}}}\sqrt{{{7}}}{\ln{{\left({x}\right)}}}\right)}}}}}{{{7}\sqrt{{{x}}}}}}\)

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