For y ≠ 0, the particular solution of the differential equation 2ye^x/ydx + (y - 2xe^x/y)dy = 0 at the point (1, 1) is

We are given:

$$2ye^{x/y}\,dx + (y - 2xe^{x/y})\,dy = 0$$2yex/ydx+(y−2xex/y)dy=0

Rearranging to express $\frac{dx}{dy}$dydx​:

$$\frac{dx}{dy} = -\frac{(y - 2xe^{x/y})}{2ye^{x/y}}$$dydx​=−2yex/y(y−2xex/y)​

$$= \frac{2xe^{x/y} - y}{2ye^{x/y}}$$=2yex/y2xex/y−y​

$$= \frac{x}{y} - \frac{1}{2e^{x/y}}$$=yx​−2ex/y1​

The repeated appearance of $x/y$x/y suggests the substitution $v = x/y$v=x/y.

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Let $v = \dfrac{x}{y}$v=yx​, so $x = vy$x=vy.

$$\frac{dx}{dy} = v + y\frac{dv}{dy}$$dydx​=v+ydydv​

Substituting into the equation:

$$v + y\frac{dv}{dy} = v - \frac{1}{2e^{v}}$$v+ydydv​=v−2ev1​

$$y\frac{dv}{dy} = -\frac{1}{2e^{v}}$$ydydv​=−2ev1​

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$$2e^{v}\,dv = -\frac{dy}{y}$$2evdv=−ydy​

$$\int 2e^{v}\,dv = -\int \frac{dy}{y}$$∫2evdv=−∫ydy​

$$2e^{v} = -\ln|y| + C$$2ev=−ln∣y∣+C

Substituting back $v = x/y$v=x/y:

$$2e^{x/y} + \ln|y| = C$$2ex/y+ln∣y∣=C

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At $x = 1$x=1, $y = 1$y=1:

$$2e^{1/1} + \ln|1| = C$$2e1/1+ln∣1∣=C

$$2e + 0 = C$$2e+0=C

$$C = 2e$$C=2e

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$$2e^{x/y} + \ln|y| = 2e$$2ex/y+ln∣y∣=2e

$$\ln|y| + 2e^{x/y} - 2e = 0$$ln∣y∣+2ex/y−2e=0

$$\ln|y| + 2(e^{x/y} - e) = 0$$ln∣y∣+2(ex/y−e)=0