What is the particular solution of the differential equation? : # dx/(x^2+x) + dy/(y^2+y) = 0 # with #y(2)=1#

The given Diff. Eqn. #dx/(x^2+x)+dy/(y^2+y)=0,# with

Initial Condition (IC) #y(2)=1.#

It is a Separable Variable Type Diff. Eqn., &, to obtain its

General Solution (GS), we integrate term-wise.

#:. intdx/{x(x+1)}+intdy/{y(y+1)}=lnc.#

#:. int{(x+1)-x}/{x(x+1)}dx+intdy/{y(y+1)}=lnc.#

#:. int{(x+1)/(x(x+1))-x/(x(x+1))}dx+intdy/{y(y+1)}=lnc.#

#:. int{1/x -1/(x+1)}dx+intdy/{y(y+1)}=lnc.#

#:.{lnx-ln(x+1)}+{lny-ln(y+1)}=lnc.#

#:. ln{x/(x+1)}+ln{y/(y+1)}=lnc.#

#:. ln{(xy)/((x+1)(y+1))}=lnc.#

#:. (xy)/((x+1)(y+1))=c," is the GS."#

To find its Particular Solution (PS), we use the given IC, that,

# y(2)=1, i.e., when, x=1, y=2.#

Subst.ing in the GS, we get, #((1)(2))/((1+1)(2+1))=c=1/3.#

This gives us the complete soln. of the eqn. :

# (xy)/((x+1)(y+1))=1/3, or, 2xy=x+y+1.#

We have an differential equation equation in the form of differentials:

# dx/(x^2+x) + dy/(y^2+y) = 0 #

We can write this in "separated variable" form as follows and integrate both sides

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

# int \ 1/(y^2+y) \ dy = -int \ 1/(x^2+x) \ dx #

Now et us find the partial fraction decomposition of #1/(u^2+u)# which we can use on both integrals:

# 1/(u^2+u) -= 1/(u(u+1)) #
# " " = A/u + B/(u+1) #
# " " = (A(u+1)+Bu)/(u(u+1)) #

Leading to:

# 1 -= A(u+1)+Bu #

We can find the constant coefficients #A# and 'B' vis substitution (effectively the "cover-up method")

Put #u=0 \ \ \ \ \ => 1=A #
Put #u=-1 => 1=-B #

Thus:

# 1/(u^2+u) -= 1/u - 1/(u+1) #

Using this in the above we get:

# int \ 1/y - 1/(y+1) \ dy = -int \ 1/x - 1/(x+1) \ dx #

We can now evaluate the integrals (not forgetting the constant of integration) to get:

# ln|y|-ln|y+1| = -{ln|x|-ln|x+1|} + c #

Then rearranging and using the properties of logarithms we have:

# ln|y|-ln|y+1| + ln|x|-ln|x+1| = c #
# :. ln ( ( |x||y|)/(|x+1| |y+1|) ) = c #
# :. ln ( |( xy)/((x+1)(y+1)) |) = c #
# :. |( xy)/((x+1)(y+1)) | = e^c #

Now #e^c > 0 AA c in RR#, thus we can remove the modulus operator, giving the General Solution:

# ( xy)/((x+1)(y+1)) = A #, say, where #A gt 0#.

We are also given that #y(2)=1#, this tells us that:

# ( 2*1)/((2+1)(1+1)) = A => A =2/3#

Thus the required solution is:

# ( xy)/((x+1)(y+1)) = 1/3 #

# :. 3xy = (x+1)(y+1) #
# :. 3xy = (xy+x+y+1) #
# :. 3xy = xy+x+y+1 #
# :. 2xy -y= x+1 #
# :. y(2x-1)= 1x+1 #
# :. y= (x+1)/(2x-1) #