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.

Question

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.

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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Methods of Solving Differential Equations> Homogeneous Differential Equations

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Q 26 Q 25.2 Q 27

2012-2013 (March) Delhi Set 1

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2012-2013 (March) Delhi Set 1 (with solutions)

Q 26 | 6 marks

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