Differential equation problem?

Question

the problem is dy/dx= y(y+1)/x

then we do this dy/y(y+1)=1/x dx

When we integrate the left side the answer is ln(y)+y. I can not figure out how to solve for y after that step?

Answer 1

move dx to the right side and divide both sides by y(y+1) so all the x's are on one side and all the y's are on the other, then integrate.

Answer 2

Your solution to the integral of dy/(y*(1+y) is not correct.  Carlos B has the correct solution (but forgot to include the constant of integration):

ln(y) - ln(y+1) = ln(x) + C

which without loss of generality can be written as

ln(y) - ln(y+1) = ln(x) + ln(a),

That is, I've defined C =  ln(a), where a is some other constant.

Now rearrange to solve for y:

ln(y) - ln(y+1) = ln(x) + ln(a),

ln(y/(y+1)) - ln(a) = ln(x)

ln(y/(a*(y+1)) = ln(x)

(y/(a*(y+1)) = x

y = (a*(y+1)) *x

y = a*y*x + a*x

y - a*x*y = a*x

y*(1 - a*x) = a*x

y = a*x/(1-a*x)

Answer 3

mmm, Partial fractions

1/y(y+1)dy=1/x dx

a/ydy + b/(y+1) dy=1/x dx

where,

a/y + b/(y+1)=1/y(y+1)

(a(y+1)+by)/y(y+1)=1/y(y+1)

a(y+1)+by=1+0y

(a+b)y+a=1+0y

Alors,

a=1

a+b=0

b=-1

Luego,

1/ydy -1/(y+1) dy=1/x dx

Integrating...

lny-ln(y+1) = lnx

Kono mama des (Something like that)