(3x^2y+e^y)dx+(x^3+xe^y-2y)dy=0 eching

$$( 3 x ^ { 2 } y + e ^ { y } ) d x + ( x ^ { 3 } + x e ^ { y } - 2 y ) d y = 0$$

d uchun yechish

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&xye^{y}+xe^{y}-2y^{2}+4yx^{3}=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

3x^{2}y+e^{y} ga d ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\left(3x^{2}yd+e^{y}d\right)x+\left(x^{3}+xe^{y}-2y\right)dy=0$$

3x^{2}yd+e^{y}d ga x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3ydx^{3}+e^{y}dx+\left(x^{3}+xe^{y}-2y\right)dy=0$$

x^{3}+xe^{y}-2y ga d ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3ydx^{3}+e^{y}dx+\left(x^{3}d+xe^{y}d-2yd\right)y=0$$

x^{3}d+xe^{y}d-2yd ga y ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3ydx^{3}+e^{y}dx+x^{3}dy+xe^{y}dy-2dy^{2}=0$$

4ydx^{3} ni olish uchun 3ydx^{3} va x^{3}dy ni birlashtirish.

$$4ydx^{3}+e^{y}dx+xe^{y}dy-2dy^{2}=0$$

d'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(4yx^{3}+e^{y}x+xe^{y}y-2y^{2}\right)d=0$$

Tenglama standart shaklda.

$$\left(xye^{y}+xe^{y}-2y^{2}+4yx^{3}\right)d=0$$

0 ni 4x^{3}y+e^{y}x+xe^{y}y-2y^{2} ga bo'lish.

$$d=0$$