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 $y={c}_{1}{y}_{1}+{c}_{2}{y}_{2}$

${x}^{2}y{}^{\u2033}-2x{y}^{\prime}+2y=0,\text{}y\left(1\right)=3,\text{}{y}^{\prime}\left(1\right)=1$

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

If r is used,

Roger Noah

Beginner2021-11-20Added 17 answers

Assume $y={x}^{r}$ then if $y={x}^{r}$ then

$y}^{\prime}=r{x}^{r-1$

And

$y{}^{\u2033}=r(r-1){x}^{r-2}$

Substitute in

${x}^{2}y{}^{\u2033}-2x{y}^{\prime}+2y=0$

to get

$r(r-1){x}^{r}-2r{x}^{r}+2{x}^{r}=0$

divide by y to get

$r(r-1)-2r+2=(r-1)(r-2)=0$

therefore

$y=ax+b{x}^{2}$

Solve

$y\left(1\right)=a+b=3$

${y}^{\prime}\left(1\right)=a+2b=1$

to get$y=5x-2{x}^{2}$

And

Substitute in

to get

divide by y to get

therefore

Solve

to get

Charles Wee

Beginner2021-11-21Added 14 answers

Assume a solution to this Euler-Cauchy equation will be proportional to

Substitute

Substitute

Assuming

The roots

Using

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My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$