1. ## Dif EQ Prob

Determine whether the points (0,1) and (2,1) are on the graph of the solution of:

(t - 1) dy + 2y dt = 0, y(0) = 1.

MY WORK:

(t - 1) dy = -2y dt
-1/(2y) dy = 1/(t - 1) dt

Take integral of both sides:

-ln(|y|)/2 = ln(|t - 1|) + C

Plug in 0 for t and plug in 1 for y:

-ln(|1|)/2 = ln(|0 - 1|) + C

=> 0 = 0 + C

=> C = 0

Now I'm not sure where to go.

Thanks for the help.

2. Originally Posted by Ideasman
Determine whether the points (0,1) and (2,1) are on the graph of the solution of:

(t - 1) dy + 2y dt = 0, y(0) = 1.

MY WORK:

(t - 1) dy = -2y dt
-1/(2y) dy = 1/(t - 1) dt

Take integral of both sides:

-ln(|y|)/2 = ln(|t - 1|) + C

Plug in 0 for t and plug in 1 for y:

-ln(|1|)/2 = ln(|0 - 1|) + C

=> 0 = 0 + C

=> C = 0

Now I'm not sure where to go.

Thanks for the help.
C=0, so,
-(1/2)ln|y| = ln|t-1|
ln|y^(-1/2)| = ln|t-1|
So,
|y^(-1/2)| = |t -1|
|1/(y^(1/2))| = |t-1|
Square both sides,
1/y = (t-1)^2 ----------(i)

Test (0,1) on (i),
1/1 =? (0-1)^2
1 =? 1
Yes, so, (0,1) is on the graph of (i).

Test (2,1) on (i),
1/1 =? (2-1)^2
1 =? 1
Yes, so, (2,1) is also on the graph of (i).