(1+y^2)dx=(tan^-1y-x)dy eching

$$( 1 + y ^ { 2 } ) d x = ( \tan ^ { - 1 } y - x ) d y$$

d uchun yechish (complex solution)

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{C}\text{, }&x=\frac{y\arctan(y)}{y^{2}+y+1}\text{ and }y\neq \frac{-1+\sqrt{3}i}{2}\text{ and }y\neq \frac{-\sqrt{3}i-1}{2}\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

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

$$\left(d+y^{2}d\right)x=\left(\arctan(y)-x\right)dy$$

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

$$dx+y^{2}dx=\left(\arctan(y)-x\right)dy$$

\arctan(y)-x ga d ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$dx+y^{2}dx=\left(\arctan(y)d-xd\right)y$$

\arctan(y)d-xd ga y ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$dx+y^{2}dx=\arctan(y)dy-xdy$$

Ikkala tarafdan \arctan(y)dy ni ayirish.

$$dx+y^{2}dx-\arctan(y)dy=-xdy$$

xdy ni ikki tarafga qo’shing.

$$dx+y^{2}dx-\arctan(y)dy+xdy=0$$

Shartlarni qayta saralash.

$$-dy\arctan(y)+dxy^{2}+dxy+dx=0$$

d'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(-y\arctan(y)+xy^{2}+xy+x\right)d=0$$

0 ni -y\arctan(y)+xy^{2}+xy+x ga bo'lish.

$$d=0$$

x uchun yechish (complex solution)

$$\left\{\begin{matrix}x=\frac{y\arctan(y)}{y^{2}+y+1}\text{, }&y\neq \frac{-1+\sqrt{3}i}{2}\text{ and }y\neq \frac{-\sqrt{3}i-1}{2}\\x\in \mathrm{C}\text{, }&d=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

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

$$\left(d+y^{2}d\right)x=\left(\arctan(y)-x\right)dy$$

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

$$dx+y^{2}dx=\left(\arctan(y)-x\right)dy$$

\arctan(y)-x ga d ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$dx+y^{2}dx=\left(\arctan(y)d-xd\right)y$$

\arctan(y)d-xd ga y ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$dx+y^{2}dx=\arctan(y)dy-xdy$$

xdy ni ikki tarafga qo’shing.

$$dx+y^{2}dx+xdy=\arctan(y)dy$$

x'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(d+y^{2}d+dy\right)x=\arctan(y)dy$$

Tenglama standart shaklda.

$$\left(dy^{2}+dy+d\right)x=dy\arctan(y)$$

Ikki tarafini d+y^{2}d+dy ga bo‘ling.

$$\frac{\left(dy^{2}+dy+d\right)x}{dy^{2}+dy+d}=\frac{dy\arctan(y)}{dy^{2}+dy+d}$$

d+y^{2}d+dy ga bo'lish d+y^{2}d+dy ga ko'paytirishni bekor qiladi.

$$x=\frac{dy\arctan(y)}{dy^{2}+dy+d}$$

\arctan(y)dy ni d+y^{2}d+dy ga bo'lish.

$$x=\frac{y\arctan(y)}{y^{2}+y+1}$$

d uchun yechish

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&x=\frac{y\arctan(y)}{y^{2}+y+1}\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

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

$$\left(d+y^{2}d\right)x=\left(\arctan(y)-x\right)dy$$

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

$$dx+y^{2}dx=\left(\arctan(y)-x\right)dy$$

\arctan(y)-x ga d ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$dx+y^{2}dx=\left(\arctan(y)d-xd\right)y$$

\arctan(y)d-xd ga y ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$dx+y^{2}dx=\arctan(y)dy-xdy$$

Ikkala tarafdan \arctan(y)dy ni ayirish.

$$dx+y^{2}dx-\arctan(y)dy=-xdy$$

xdy ni ikki tarafga qo’shing.

$$dx+y^{2}dx-\arctan(y)dy+xdy=0$$

Shartlarni qayta saralash.

$$-dy\arctan(y)+dxy^{2}+dxy+dx=0$$

d'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(-y\arctan(y)+xy^{2}+xy+x\right)d=0$$

0 ni -y\arctan(y)+xy^{2}+xy+x ga bo'lish.

$$d=0$$

x uchun yechish

$$\left\{\begin{matrix}\\x=\frac{y\arctan(y)}{y^{2}+y+1}\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&d=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

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

$$\left(d+y^{2}d\right)x=\left(\arctan(y)-x\right)dy$$

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

$$dx+y^{2}dx=\left(\arctan(y)-x\right)dy$$

\arctan(y)-x ga d ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$dx+y^{2}dx=\left(\arctan(y)d-xd\right)y$$

\arctan(y)d-xd ga y ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$dx+y^{2}dx=\arctan(y)dy-xdy$$

xdy ni ikki tarafga qo’shing.

$$dx+y^{2}dx+xdy=\arctan(y)dy$$

x'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(d+y^{2}d+dy\right)x=\arctan(y)dy$$

Tenglama standart shaklda.

$$\left(dy^{2}+dy+d\right)x=dy\arctan(y)$$

Ikki tarafini d+y^{2}d+dy ga bo‘ling.

$$\frac{\left(dy^{2}+dy+d\right)x}{dy^{2}+dy+d}=\frac{dy\arctan(y)}{dy^{2}+dy+d}$$

d+y^{2}d+dy ga bo'lish d+y^{2}d+dy ga ko'paytirishni bekor qiladi.

$$x=\frac{dy\arctan(y)}{dy^{2}+dy+d}$$

\arctan(y)dy ni d+y^{2}d+dy ga bo'lish.

$$x=\frac{y\arctan(y)}{y^{2}+y+1}$$