Differential equations: separating variables

[armolinasf]

Homework Statement

So I have two problems that ask me to separate variables:

1: du/dx * x(x+1) = u^2 u(1)=1

2: dy/dt=y(2-y) Initial condition y(0)=1

The Attempt at a Solution

For #1 i get each variable on one side and have integral(u^-2du)=integral(dx/x(x+1)
I use partial fractions to break up the x side, coefficient are 1 and -1. Now I have:
-1/u=ln(x/x+1) +C ==> u=-1/(ln(x)-ln(x+1)) + C. And evaluated at (1,1) C=ln2.

that would give us, u=-1/(ln(x)-ln(x+1)) + ln2, however the answer in my book gives u=-1/((ln2x/x+1)-1) I'm having trouble seeing where the minus one in the denominator comes from can someone show me an explanation?

For #2 i get integral(dy/y(2-y)) = integral(dt) I again use partial fractions to break up the y side, giving coefficients of 1/2. I'm left with 1/2(ln(y) + ln(2-y))=t. I'm having trouble isolating y on one side. I need a point in the right direction.

Thanks for the help!

 

[CompuChip]

For both questions, playing around with the identities
ln(a) + ln(b) = ln(ab)
ln(a) - ln(b) = ln(a/b)
may be useful.

 

[magicarpet512]

u=-1/(ln(x)-ln(x+1)) + C. And evaluated at (1,1) C=ln2.

I don't think this is true... by properties of logarithms we have
u = -1/(ln(x/x+1)) + C.
Evaluated at (1,1) C = 1 + 1/ln(1/2)

 

[Mark44]

Homework Statement

So I have two problems that ask me to separate variables:

1: du/dx * x(x+1) = u^2 u(1)=1

2: dy/dt=y(2-y) Initial condition y(0)=1

The Attempt at a Solution

For #1 i get each variable on one side and have integral(u^-2du)=integral(dx/x(x+1)
I use partial fractions to break up the x side, coefficient are 1 and -1. Now I have:
-1/u=ln(x/x+1) +C ==> u=-1/(ln(x)-ln(x+1)) + C. And evaluated at (1,1) C=ln2.

that would give us, u=-1/(ln(x)-ln(x+1)) + ln2, however the answer in my book gives u=-1/((ln2x/x+1)-1) I'm having trouble seeing where the minus one in the denominator comes from can someone show me an explanation?

For #2 i get integral(dy/y(2-y)) = integral(dt) I again use partial fractions to break up the y side, giving coefficients of 1/2. I'm left with 1/2(ln(y) + ln(2-y))=t. I'm having trouble isolating y on one side. I need a point in the right direction.

Thanks for the help!

I didn't work #1, but maybe my help with #2 will help you with that one.

You have a sign error when you integrated the left side.
1/2 \int [1/y + 1/(2 - y)] dy = \int dt
\Rightarrow 1/2 [ ln|y| - ln|2 - y| ]= t + C

In the second integral on the left side you have \int dy/(2 - y). Use an ordinary substitution to evaluate this integral.

Since y(0) = 1, then C = 0.

Now, since y(0) = 1, when t is near 0, y is near 1, so that should tell you something about whether you can get rid of the absolute values for |y| and |2 - y|.