By using the substitution y=vx, find the particular solution of the differential equation 2xy (dy/dx)=y^2-x^2 when y=4 and x=2. Express y in terms of x.?

1 Answer
Ananda Dasgupta
Mar 15, 2018

y = x sqrt(10/x-1)

Explanation:

We use the substitution y=vx. This means that

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

Now, the given equation is

2xy (dy/dx)=y^2-x^2
or
dy/dx=1/2(y/x-x/y)

so that

v+x (dv)/dx =1/2(v-1/v) implies
x (dv)/dx =-1/2(1/v+v)
which can be rewritten in the form

2 v/(v^2+1) dv=-(dx)/x implies

(d(v^2+1))/(v^2+1) =-dx/x

which readily integrates to

ln(v^2+1) = -ln x + C

The given initial condition is y=4 at x=2. This means that at x=2, v=4/2=2. Substituting this in the solution we have obtained so far, we get

ln 5=-ln 2 +C

giving the value of the constant of integration as ln 10.

The solution is thus

(v^2+1) = 10/x

or

y^2/x^2+1 =10/x

or

y = x sqrt(10/x-1)

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