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

Show that the differential equation 2yx/y dx + (y − 2x ex/y) dy = 0 is homogeneous. Find the particular solution of this differential equation, given that x = 0 when y = 1.

#### Solution

The given differential equation can be written as

dx/dy=(2xe^(x/y)-y)/(2ye^(x/y)) .....................(1)

Let F(x,y)=(2xe^(x/y)-y)/(2ye^(x/y))

then  F(lambdax,lambday)=(lambda(2xe^(x/y)-y))/(lambda(2ye^(x/y)))=lambda^@[F(x,y)]

Thus, F (x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.

For solving, let us substitute x=vy     ..................(2)

Differentiating equation (2) with respect to y, we get

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

Substituting the value of x and  dx/dy  in equation (1), we get

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

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

or y(dv)/(dy)=-1/(2e^v)

or 2e^vdv=(-dy)/y

or int 2e^vdv=-intdy/y

or 2e^v=-log|y|+C

Replacing v by x/y , we get

2e^(x/y)+log|y|=c......(3)

Substituting x = 0 and y = 1in equation (3), we get

2e^0+log|1|=c =>c=2

Substituting the value of C in equation (3), we get

2e^(x/y)+log|y|=2 .which is the particular solution of the given differential equation.

Concept: Methods of Solving First Order, First Degree Differential Equations - Homogeneous Differential Equations
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