Kan noen plz forklare ?( matte R2)

[2fast4you]

I boka er det et eks. der de viser en overgang fra (1)-(2) men skjønner ikke helt denne overgangen, kan noen plz forklare det ?

(1) (u``+2ru`+r^2u)e^rx - 6(u`+ru)e^rx + 5u*e^rx = 0

vi dividerer med e^rx. Ordner likningen og får

(2) u``+ (2r-6)*u`+ (r^2-6r+5)*u = 0

takk!

[Nebuchadnezzar]

[tex]\left( {u^{\tiny\prime\prime} + 2ru^{\tiny\prime} + {r^2}u} \right){e^rx} - 6\left( {u^{\tiny\prime} + ru} \right){e^{rx}} + 5u \cdot {e^{rx}} = 0 [/tex]

[tex] \frac{{\left( {u^{\tiny\prime\prime} + 2ru^{\tiny\prime} + {r^2}u} \right){e^rx} - 6\left( {u^{\tiny\prime} + ru} \right){e^{rx}} + 5u \cdot {e^{rx}}}}{{{e^r}x}} = \frac{0}{{{e^r}x}} [/tex]

[tex] \frac{{\left( {u^{\tiny\prime\prime} + 2ru^{\tiny\prime} + {r^2}u} \right){e^rx}}}{{{e^r}x}} - \frac{{6\left( {u^{\tiny\prime} + ru} \right){e^{rx}}}}{{{e^r}x}} + \frac{{5u \cdot {e^{rx}}}}{{{e^r}x}} = \frac{0}{{{e^r}x}} [/tex]

[tex] u^{\tiny\prime\prime} + 2ru^{\tiny\prime} + {r^2}u - 6u^{\tiny\prime} - 6ru + 5u = 0 [/tex]

[tex] u^{\tiny\prime\prime} + 2ru^{\tiny\prime} - 6u^{\tiny\prime} - 6ru + 5u + {r^2}u = 0 [/tex]

[tex] u^{\tiny\prime\prime} + \left( {2r - 6} \right)u^{\tiny\prime} + \left( {{r^2} - 6r + 5} \right)u = 0 [/tex]

[2fast4you]

takk