Whakaoti i te dy/dydydx=1-1/sqrt{2}

$$\frac { d y } { d y } d y d x = 1 - \frac { 1 } { \sqrt { 2 } }$$

Whakaoti mō x

$$x=\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{2yd^{2}}$$

$$d\neq 0\text{ and }y\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakareatia te d ki te d, ka d^{2}.

$$\frac{\mathrm{d}(y)}{\mathrm{d}y}d^{2}yx=1-\frac{1}{\sqrt{2}}$$

Whakangāwaritia te tauraro o \frac{1}{\sqrt{2}} mā te whakarea i te taurunga me te tauraro ki te \sqrt{2}.

$$\frac{\mathrm{d}(y)}{\mathrm{d}y}d^{2}yx=1-\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}$$

Ko te pūrua o \sqrt{2} ko 2.

$$\frac{\mathrm{d}(y)}{\mathrm{d}y}d^{2}yx=1-\frac{\sqrt{2}}{2}$$

Whakareatia ngā taha e rua o te whārite ki te 2.

$$2\frac{\mathrm{d}(y)}{\mathrm{d}y}d^{2}yx=2-\sqrt{2}$$

He hanga arowhānui tō te whārite.

$$2yd^{2}x=2-\sqrt{2}$$

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

$$\frac{2yd^{2}x}{2yd^{2}}=\frac{2-\sqrt{2}}{2yd^{2}}$$

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

$$x=\frac{2-\sqrt{2}}{2yd^{2}}$$

Whakaoti mō d (complex solution)

$$d=-\frac{\sqrt{4-2\sqrt{2}}x^{-\frac{1}{2}}y^{-\frac{1}{2}}}{2}$$

$$d=\frac{\sqrt{4-2\sqrt{2}}x^{-\frac{1}{2}}y^{-\frac{1}{2}}}{2}\text{, }x\neq 0\text{ and }y\neq 0$$

Whakaoti mō d

$$d=\frac{\sqrt{\frac{4-2\sqrt{2}}{xy}}}{2}$$

$$d=-\frac{\sqrt{\frac{4-2\sqrt{2}}{xy}}}{2}\text{, }\left(x>0\text{ and }y>0\right)\text{ or }\left(y<0\text{ and }x<0\right)$$