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

##### 2 Answers

#### Explanation:

The given **Diff. Eqn.**

**Initial Condition (IC)**

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

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

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

Subst.ing in the **GS,** we get,

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

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

#### Explanation:

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) -= 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

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

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

We are also given 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) #