Whakaoti i te 0quad(1+y^2)dx=(tan^-1y^prime-x)dy

$$0 \quad ( 1 + y ^ { 2 } ) d x = ( \tan ^ { - 1 } y ^ { \prime } - x ) d y$$

Whakaoti mō d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&x=0\text{ or }y=0\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Ko te tau i whakarea ki te kore ka hua ko te kore.

$$0=\left(\arctan(\frac{\mathrm{d}}{\mathrm{d}x}(y))-x\right)dy$$

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

$$0=\left(\arctan(\frac{\mathrm{d}}{\mathrm{d}x}(y))d-xd\right)y$$

Whakamahia te āhuatanga tohatoha hei whakarea te \arctan(\frac{\mathrm{d}}{\mathrm{d}x}(y))d-xd ki te y.

$$0=\arctan(\frac{\mathrm{d}}{\mathrm{d}x}(y))dy-xdy$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$\arctan(\frac{\mathrm{d}}{\mathrm{d}x}(y))dy-xdy=0$$

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

$$\left(\arctan(\frac{\mathrm{d}}{\mathrm{d}x}(y))y-xy\right)d=0$$

He hanga arowhānui tō te whārite.

$$\left(-xy+\arctan(0)y\right)d=0$$

Whakawehe 0 ki te \arctan(0)y-xy.

$$d=0$$

Whakaoti mō x

$$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&d=0\text{ or }y=0\end{matrix}\right.$$