# Solve the differential equation # (2y-x)dy/dx=2x+y# where #y=3# when #x=2#?

##### 2 Answers

See below.

#### Explanation:

Making

and simplifying

For initial conditions we have

# y^2-xy-x^2+1=0#

#### Explanation:

We have:

# (2y-x)dy/dx=2x+y #

We can rearrange this Differential Equation as follows:

# dy/dx = (2x+y)/(2y-x) #

# " " = (2x+y)/(2y-x) * (1/x)/(1/x) #

# " " = (2+(y/x))/(2(y/x)-1) #

So Let us try a substitution, Let:

# v = y/x => y=vx#

Then:

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

And substituting into the above DE, to eliminate

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

# :. x(dv)/dx = (2+v)/(2v-1) - v#

# :. " " = {(2+v) - v(2v-1)}/(2v-1) #

# :. " " = {2+v - 2v^2+v}/(2v-1)#

# :. " " = {2+2v - 2v^2}/(2v-1)#

# :. " " = -(2(v^2-v-1))/(2v-1)#

# :. (2v-1)/(v^2-v-1) \ (dv)/dx = -2/x #

This is now a separable DIfferential Equation, and so "separating the variables" gives us:

# int \ (2v-1)/(v^2-v-1) \ dv = - int \ 2/x \ dx #

This is now a trivial integration problem, thus:

# ln(v^2-v-1) = -2lnx + lnA #

# " " = lnA/x^2 #

# :. v^2-v-1 = A/x^2 #

And restoring the substitution we get:

# (y/x)^2-(y/x)-1 = A/x^2 #

Using

# 9/4-3/2-1 = A/4 => A = -1#

Thus:

# (y/x)^2-(y/x)-1 = -1/x^2#

# :. y^2-xy-x^2=-1#

# :. y^2-xy-x^2+1=0#