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

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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Roger Noah
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

$$\displaystyle{y}{''}{\left({x}\right)}{x}^{{2}}+{2}{y}'{\left({x}\right)}{x}+{2}{y}{\left({x}\right)}={0}$$
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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