- #1
aerograce
- 64
- 1
Let y be a solution of the di®erential equation,
[itex]\frac{dy}{dx}[/itex]=[itex]\frac{y-1}{x+1}[/itex]
such that
y(0)=0;
Then y(1) is:
My attempt:
[itex]\frac{1}{y-1}[/itex]dy=[itex]\frac{1}{x+1}[/itex]dx
ln|y-1|= ln |x+1| +C
Problems start to occur from here:
My method:
When x=0, y=0. Hence ln0=ln0+C C=0;
When x=1, ln|y-1|= ln2 + 0;
Hence |y-1| = 2
y=3 or y=-1.
My teacher's method
[itex]\frac{1}{y-1}[/itex]dy=[itex]\frac{1}{x+1}[/itex]dx
ln|y-1|= ln |x+1| +C
(y-1)=A(x+1)
Because y(0)=0 Hence A=-1
And when x=1, y-1=-2 y=-1
I am quite confused. My method seems to be legit also, but why I got different answer.
[itex]\frac{dy}{dx}[/itex]=[itex]\frac{y-1}{x+1}[/itex]
such that
y(0)=0;
Then y(1) is:
My attempt:
[itex]\frac{1}{y-1}[/itex]dy=[itex]\frac{1}{x+1}[/itex]dx
ln|y-1|= ln |x+1| +C
Problems start to occur from here:
My method:
When x=0, y=0. Hence ln0=ln0+C C=0;
When x=1, ln|y-1|= ln2 + 0;
Hence |y-1| = 2
y=3 or y=-1.
My teacher's method
[itex]\frac{1}{y-1}[/itex]dy=[itex]\frac{1}{x+1}[/itex]dx
ln|y-1|= ln |x+1| +C
(y-1)=A(x+1)
Because y(0)=0 Hence A=-1
And when x=1, y-1=-2 y=-1
I am quite confused. My method seems to be legit also, but why I got different answer.