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

Answer 1

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

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

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

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.#
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Answer 3

The particular solution of the given differential equation is ( y(x) = \frac{1}{x+1} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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