Whakaoti i te (1+y^2)dx=(tan^-1y-x)dy

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

Whakaoti mō d

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

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te 1+y^{2} ki te d.

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

Whakamahia te āhuatanga tohatoha hei whakarea te d+y^{2}d ki te x.

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

Whakamahia te āhuatanga tohatoha hei whakarea te \arctan(y)-x ki te d.

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

Whakamahia te āhuatanga tohatoha hei whakarea te \arctan(y)d-xd ki te y.

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

Tangohia te \arctan(y)dy mai i ngā taha e rua.

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

Me tāpiri te xdy ki ngā taha e rua.

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

Whakaraupapatia anō ngā kīanga tau.

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

Pahekotia ngā kīanga tau katoa e whai ana i te d.

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

Whakawehe 0 ki te -y\arctan(y)+xy^{2}+xy+x.

$$d=0$$

Whakaoti mō x

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

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te 1+y^{2} ki te d.

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

Whakamahia te āhuatanga tohatoha hei whakarea te d+y^{2}d ki te x.

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

Whakamahia te āhuatanga tohatoha hei whakarea te \arctan(y)-x ki te d.

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

Whakamahia te āhuatanga tohatoha hei whakarea te \arctan(y)d-xd ki te y.

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

Me tāpiri te xdy ki ngā taha e rua.

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

Pahekotia ngā kīanga tau katoa e whai ana i te x.

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

He hanga arowhānui tō te whārite.

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

Whakawehea ngā taha e rua ki te d+y^{2}d+dy.

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

Mā te whakawehe ki te d+y^{2}d+dy ka wetekia te whakareanga ki te d+y^{2}d+dy.

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

Whakawehe \arctan(y)dy ki te d+y^{2}d+dy.

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