Y(s)/(x_{s)}=1/s(s+1)(s+2) eching

$$\frac { Y ( s ) } { ( x _ { s } ) } = \frac { 1 } { s ( s + 1 ) ( s + 2 ) }$$

Y uchun yechish

$$Y=\frac{x_{s}}{\left(s+1\right)\left(s+2\right)s^{2}}$$

$$x_{s}\neq 0\text{ and }s\neq 0\text{ and }s\neq -1\text{ and }s\neq -2$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglamaning ikkala tarafini sx_{s}\left(s+1\right)\left(s+2\right) ga, x_{s},s\left(s+1\right)\left(s+2\right) ning eng kichik karralisiga ko‘paytiring.

$$s\left(s+1\right)\left(s+2\right)Ys=x_{s}$$

s ga s+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\left(s^{2}+s\right)\left(s+2\right)Ys=x_{s}$$

s^{2}+s ga s+2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$\left(s^{3}+3s^{2}+2s\right)Ys=x_{s}$$

s^{3}+3s^{2}+2s ga Y ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\left(s^{3}Y+3s^{2}Y+2sY\right)s=x_{s}$$

s^{3}Y+3s^{2}Y+2sY ga s ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$Ys^{4}+3Ys^{3}+2Ys^{2}=x_{s}$$

Y'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(s^{4}+3s^{3}+2s^{2}\right)Y=x_{s}$$

Ikki tarafini s^{4}+3s^{3}+2s^{2} ga bo‘ling.

$$\frac{\left(s^{4}+3s^{3}+2s^{2}\right)Y}{s^{4}+3s^{3}+2s^{2}}=\frac{x_{s}}{s^{4}+3s^{3}+2s^{2}}$$

s^{4}+3s^{3}+2s^{2} ga bo'lish s^{4}+3s^{3}+2s^{2} ga ko'paytirishni bekor qiladi.

$$Y=\frac{x_{s}}{s^{4}+3s^{3}+2s^{2}}$$

x_{s} ni s^{4}+3s^{3}+2s^{2} ga bo'lish.

$$Y=\frac{x_{s}}{\left(s+1\right)\left(s+2\right)s^{2}}$$