# Find the particular solution of the differential equation 2y e^(x/y) dx + (y - 2x e^(x/y)) dy = 0 given that x = 0 when y = 1. - Mathematics

Find the particular solution of the differential equation

2y ex/y dx + (y - 2x ex/y) dy = 0

given that x = 0 when y = 1.

#### Solution

2ye^(x/y)dx+(y-2xe^(x/y))dy=0

=>dx/dy=(2xe^(x/y-y))/(2ye^(x/y))

Given differential equation is a homogeneous differential equation.

∴ Put x = vy

dx/dy=v+y (dv)/dy

v+y(dv)/dy=(2ve^v-1)/(2e^v)

=>y(dv)/dy=(2ve^v-1)/(2e^v)-v

=>y(dv)/dy=-1/(2e^v)

=>2e^vdv=-1/ydy

Integrating on both the sides

=>2inte^vdv=-int1/ydy

=>2e^v=-log|y|+logC

=>2e^v=log|c/y|

=>2e^(x/y)=log|c/y|

Given that at x = 0, y = 1

2e^0= log|c/1|

⇒ C = e2

:.2e^(x/y)=log""e^2/y

=>logy=-2e^(x/y)+2

=>y=e^2-2e^(x/y)

Concept: General and Particular Solutions of a Differential Equation
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