User:Egm6321.f11.team2.Xia/RP6.6

=R*6.6 - Finding L2-ODE-VC =

Given
Trial solution

$$\displaystyle y(x) = \frac {e^{rx}}{\sin x}$$ (6.6.1)

and characteristic equation.

$$\displaystyle (r-2)(r-\frac {1}{1+x})=0$$ (6.6.2)

Find
Find an L2-ODE-VC whose characteristic equation is 6.6.2

Solution
From 6.6.1, we have

$$\displaystyle y'= \frac{r\,{e}^{r\,x}}{\mathrm{sin}\left( x\right) }-\frac{{e}^{r\,x}\,\mathrm{cos}\left( x\right) }{{\mathrm{sin}\left( x\right) }^{2}}$$ (6.6.3)

$$\displaystyle y''= \frac{{r}^{2}\,{e}^{r\,x}}{\mathrm{sin}\left( x\right) }+\frac{{e}^{r\,x}}{\mathrm{sin}\left( x\right) }-\frac{2\,r\,{e}^{r\,x}\,\mathrm{cos}\left( x\right) }{{\mathrm{sin}\left( x\right) }^{2}}+\frac{2\,{e}^{r\,x}\,{\mathrm{cos}\left( x\right) }^{2}}{{\mathrm{sin}\left( x\right) }^{3}}$$ (6.6.4) then

$$\displaystyle y'=\left(r-\frac{\,\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}\right)\frac{{e}^{r\, x}}{\mathrm{sin}\left(x\right)}$$ (6.6.5)

$$\displaystyle y''=\left({r}^{2}+1-\frac{2\, r\,\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}+\frac{2\,{\mathrm{cos}\left(x\right)}^{2}}{{\mathrm{sin}\left(x\right)}^{2}}\right)\frac{{e}^{r\, x}}{\mathrm{sin}\left(x\right)}$$ (6.6.6) So we can assume that

$$\displaystyle a_{2}(x)\left({r}^{2}+1-\frac{2\, r\,\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}+\frac{2\,{\mathrm{cos}\left(x\right)}^{2}}{{\mathrm{sin}\left(x\right)}^{2}}\right)\frac{{e}^{r\, x}}{\mathrm{sin}\left(x\right)}+a_{1}(x)\left(r-\frac{\,\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}\right)\frac{{e}^{r\, x}}{\mathrm{sin}\left(x\right)}+a_{0}(x)\frac{{e}^{r\, x}}{\mathrm{sin}\left(x\right)}=0$$ (6.6.7)

$$\displaystyle a_{2}(x){r}^{2}+\left(a_{1}(x)-a_{2}(x)\frac{2\,\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}\right)r-a_{1}(x)\frac{cos\left(x\right)}{\mathrm{sin}\left(x\right)}+a_{2}(x)+a_{2}(x)\frac{2\,{\mathrm{cos}\left(x\right)}^{2}}{{\mathrm{sin}\left(x\right)}^{2}}+a_{0}(x)=0$$ (6.6.8) comparing 6.6.8 with the 6.6.2 we have

$$\displaystyle a_{2}=x+1$$ (6.6.9) $$\displaystyle a_{1}=2(x+1)\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}-(2x-3)$$ (6.6.10) $$\displaystyle a_{0}=1-x-(2x-3)\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}$$ (6.6.11) Thus

$$\displaystyle \left(x+1\right)y''+\left(2(x+1)\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}-(2x-3)\right)y'+\left(1-x-(2x-3)\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}\right)y=0$$ (6.6.12)