Question

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

Answer 1

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

`(2) : 2xy=x+y+1," is the PS."`

Explanation:

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.`

Answer 2

`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)` 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)`