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.?

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Answer 1 · 2018-03-15T11:16:36

#y = x sqrt(10/x-1)#

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)#