L_{f}=L_{0}(1+alphaDeltat) eching

$$L _ { f } = L _ { 0 } ( 1 + \alpha \Delta t )$$

L_0 uchun yechish

$$\left\{\begin{matrix}L_{0}=\frac{L_{f}}{t\Delta \alpha +1}\text{, }&t=0\text{ or }\Delta =0\text{ or }\alpha \neq -\frac{1}{t\Delta }\\L_{0}\in \mathrm{R}\text{, }&L_{f}=0\text{ and }\alpha =-\frac{1}{t\Delta }\text{ and }t\neq 0\text{ and }\Delta \neq 0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

L_{0} ga 1+\alpha \Delta t ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$L_{f}=L_{0}+L_{0}\alpha \Delta t$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$L_{0}+L_{0}\alpha \Delta t=L_{f}$$

L_{0}'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(1+\alpha \Delta t\right)L_{0}=L_{f}$$

Tenglama standart shaklda.

$$\left(t\Delta \alpha +1\right)L_{0}=L_{f}$$

Ikki tarafini 1+\alpha \Delta t ga bo‘ling.

$$\frac{\left(t\Delta \alpha +1\right)L_{0}}{t\Delta \alpha +1}=\frac{L_{f}}{t\Delta \alpha +1}$$

1+\alpha \Delta t ga bo'lish 1+\alpha \Delta t ga ko'paytirishni bekor qiladi.

$$L_{0}=\frac{L_{f}}{t\Delta \alpha +1}$$

L_f uchun yechish

$$L_{f}=L_{0}\left(t\Delta \alpha +1\right)$$