I am a self studier, there might be things seems obvious to you guys that are not obvious for me. I have a few questions here, I write down the equation first and present the question in the last part. Please bare with me. Thanks.(adsbygoogle = window.adsbygoogle || []).push({});

Eulers equation of 2nd order: [tex]L[y]=x^{2} y''+ \alpha xy'+ \beta y = 0[/tex] (3)

Let [tex]y = x^{r}[/tex]

=> [tex]L[x^{r}]=x^{r}[r(r-1)+\alpha x(x^{r})'+ \beta (x^{r})] = 0[/tex]

If r is a root of the quadratic equation: [tex]F(r)=r(r-1)+\alpha r + \beta = 0[/tex]

[tex] r_{1} , r_{2} =\frac{-(\alpha - 1)^{+}_{-}\sqrt{(\alpha-1)^{2}-4\beta}}{2}[/tex]

and

[tex]F(r)=(r-r_{1})(r-r_{2})[/tex]

For [tex]r_{1} = r_{2}[/tex], [tex]F(r)=(r-r_{1})^{2}[/tex], Both F(r) and F'(r) =0.

We only find [tex]y_{1}=x^{r}[/tex]. We need to find [tex]y_{2}.[/tex]

We differentiating(3) [tex]\frac{d}{dr}[L(x^{r}]=\frac{d}{dr}{x^{r}F(r)][/tex](4)

[tex]\frac{dx^{r}}{dr}=x^{r}ln(x)[/tex]

[tex]L[x^{r}ln(x)]=(r-r_{1})^{2}x^{r}ln(x)+2(r-r_{1})x^{r}[/tex](5)

For [tex]r=r_{1}[/tex], right side of (5) equal to zero.

Therefore [tex]y_{2}(x)=x^{r_{1}}ln(x).[/tex]

My questions are:

1) How is [tex]\frac{d}{dr}[L(x^{r}]=\frac{d[x^{r}ln(x)]}{dr}[/tex] from (4) to (5)?

2) Why For [tex]r=r_{1}[/tex], right side of (5) equal to zero. And therefore [tex]y_{2}(x)=x^{r_{1}}ln(x)?[/tex]

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# Please help with Euler DE.

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