protisvitfc

2021-11-19

I'm trying to find particular solution of the second-order linear equation but I can't find ${y}_{1}$ and ${y}_{2}$ according to $y={c}_{1}{y}_{1}+{c}_{2}{y}_{2}$

If r is used, ${x}^{2}{r}^{2}-2xr=-2$ then $xr\left(xr-2\right)=-2$, I can't go on from there to find ${y}_{1}$ and ${y}_{2}$

Roger Noah

Assume $y={x}^{r}$ then if $y={x}^{r}$ then
${y}^{\prime }=r{x}^{r-1}$
And
$y{}^{″}=r\left(r-1\right){x}^{r-2}$
Substitute in
${x}^{2}y{}^{″}-2x{y}^{\prime }+2y=0$
to get
$r\left(r-1\right){x}^{r}-2r{x}^{r}+2{x}^{r}=0$
divide by y to get
$r\left(r-1\right)-2r+2=\left(r-1\right)\left(r-2\right)=0$
therefore
$y=ax+b{x}^{2}$
Solve
$y\left(1\right)=a+b=3$
${y}^{\prime }\left(1\right)=a+2b=1$
to get $y=5x-2{x}^{2}$

Charles Wee

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

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