So, I apparently you put the y on the other side and integrate separately: INT 1/y day = INT 1/[x(x+1)] dx ...
However, I want to know why we can't flip the other part: INT y dy = INT x(x+1) dx ???
Solve the differential equation: y/ [x(x+1)]?
2016-09-05 7:54 pm
回答 (2)
2016-09-05 8:40 pm
If you flip it, then it will be y/dy = x (x+1)/dx. You are not going to be better.
2016-09-05 7:59 pm
Why would you?
dy/dx = y / (x * (x + 1))
That's your starting point
dy / y = dx / (x * (x + 1))
There's nothing new here. These are the same old fractions you have been working with since the 2nd grade. You're not "putting" anything anywhere. You're doing what pretty much amounts to cross-multiplication
A/x + B/(x + 1) = 1/(x * (x + 1))
A * (x + 1) + B * x = 0x + 1
Ax + Bx + A = 0x + 1
A + B = 0
A = 1
B = -1
dy/y = dx/x - dx/(x + 1)
Integrate
ln|y| = ln|x| - ln|x + 1| + C
ln|y| = ln|x/(x + 1)| + C
y = C * (x/(x + 1))
y = Cx / (x + 1)
dy/dx = y / (x * (x + 1))
That's your starting point
dy / y = dx / (x * (x + 1))
There's nothing new here. These are the same old fractions you have been working with since the 2nd grade. You're not "putting" anything anywhere. You're doing what pretty much amounts to cross-multiplication
A/x + B/(x + 1) = 1/(x * (x + 1))
A * (x + 1) + B * x = 0x + 1
Ax + Bx + A = 0x + 1
A + B = 0
A = 1
B = -1
dy/y = dx/x - dx/(x + 1)
Integrate
ln|y| = ln|x| - ln|x + 1| + C
ln|y| = ln|x/(x + 1)| + C
y = C * (x/(x + 1))
y = Cx / (x + 1)
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