Resolver V_{12}=v_{12}(-i(phi_{12}+p_{1}-p_{2}))

$$V _ { 12 } = v _ { 12 } ( - i ( \phi _ { 12 } + p _ { 1 } - p _ { 2 } ) )$$

Ebatzi: p_1

$$\left\{\begin{matrix}p_{1}=p_{2}-ϕ_{12}+\frac{iV_{12}}{v_{12}}\text{, }&v_{12}\neq 0\\p_{1}\in \mathrm{C}\text{, }&V_{12}=0\text{ and }v_{12}=0\end{matrix}\right.$$

Ekuazio lineala ebazteko urratsak

Erabili banaketa-propietatea v_{12}\left(-i\right) eta ϕ_{12}+p_{1}-p_{2} biderkatzeko.

$$V_{12}=-iv_{12}ϕ_{12}-iv_{12}p_{1}+iv_{12}p_{2}$$

Trukatu aldeak, aldagaiak ezkerraldean egon daitezen.

$$-iv_{12}ϕ_{12}-iv_{12}p_{1}+iv_{12}p_{2}=V_{12}$$

Kendu -iv_{12}ϕ_{12} bi aldeetatik.

$$-iv_{12}p_{1}+iv_{12}p_{2}=V_{12}-\left(-iv_{12}ϕ_{12}\right)$$

Kendu iv_{12}p_{2} bi aldeetatik.

$$-iv_{12}p_{1}=V_{12}-\left(-iv_{12}ϕ_{12}\right)-iv_{12}p_{2}$$

i lortzeko, biderkatu -1 eta -i.

$$-iv_{12}p_{1}=V_{12}+iv_{12}ϕ_{12}-iv_{12}p_{2}$$

Modu arruntean dago ekuazioa.

$$\left(-iv_{12}\right)p_{1}=V_{12}+iv_{12}ϕ_{12}-ip_{2}v_{12}$$

Zatitu ekuazioaren bi aldeak -iv_{12} balioarekin.

$$\frac{\left(-iv_{12}\right)p_{1}}{-iv_{12}}=\frac{V_{12}+iv_{12}ϕ_{12}-ip_{2}v_{12}}{-iv_{12}}$$

-iv_{12} balioarekin zatituz gero, -iv_{12} balioarekiko biderketa desegiten da.

$$p_{1}=\frac{V_{12}+iv_{12}ϕ_{12}-ip_{2}v_{12}}{-iv_{12}}$$

Zatitu V_{12}+iv_{12}ϕ_{12}-iv_{12}p_{2} balioa -iv_{12} balioarekin.

$$p_{1}=p_{2}-ϕ_{12}+\frac{iV_{12}}{v_{12}}$$

Ebatzi: V_12

$$V_{12}=-iv_{12}\left(p_{1}-p_{2}+ϕ_{12}\right)$$