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Question
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.
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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.
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