dy/dydydx=1-1/sqrt{2} eching

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

x uchun yechish

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

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

Chiziqli tenglamalarni yechish bo'yicha qadamlar

d^{2} hosil qilish uchun d va d ni ko'paytirish.

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

\frac{1}{\sqrt{2}} maxrajini \sqrt{2} orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.

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

\sqrt{2} kvadrati – 2.

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

Tenglamaning ikkala tarafini 2 ga ko'paytirish.

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

Tenglama standart shaklda.

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

Ikki tarafini 2d^{2}y ga bo‘ling.

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

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

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

d uchun yechish (complex solution)

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

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

d uchun yechish

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

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