# Thread: Help with variables seperable ODEs needed

1. ## Help with variables seperable ODEs needed

Ok, so i just started a 2nd year engineering maths unit, but i got credit for the 1st year prerequisite from a subject i did about 4 yrs ago...
need some help.

eg. dy/dt + 2y = 1 , y(0)= 5/2

i started with dy/(1-2y)= dt
ln(1-2y) = t+c
1-2y = e^(t+c)
y= 1/2(1-e^(t+c))
am i right so far?

i then put y(0)= 1/2(1-e^c) = 5/2
1-e^c = 5
e^c = -4
c = ln(-4) which is clearly wrong.
where did i go wrong? any help greatly appreciated.. thanks

2. Originally Posted by linalg123
i started with dy/(1-2y)= dt ln(1-2y) = t+c

It should be $-(1/2)\ln |1-2y|=t+c$ .

3. ok, so -(1/2)ln(1-2y) = t+c
ln(1-2y) = -2t - 2c
1-2y = e^(-2t-2c)
y= 1/2( 1 - e(-2t-2c))
is this right?
because i am still getting a c=ln(negative) when i try to solve the i.v.p

4. Originally Posted by linalg123
ok, so -(1/2)ln(1-2y) = t+c
ln(1-2y) = -2t - 2c
1-2y = e^(-2t-2c)
y= 1/2( 1 - e(-2t-2c))
This can reduce to $\displaystyle y(t)=\frac{1}{2}(1-Ae^{-2t})$

where A=-2c.

Now continue on and find A using your condition.

5. can you explain the step how 1/2(1- e(-2t-2c)) = 1/2 (1- Ae^-2t) where A= -2c.. wouldn't A= e^-2c?

6. Originally Posted by linalg123
can you explain the step how 1/2(1- e(-2t-2c)) = 1/2 (1- Ae^-2t) where A= -2c.. wouldn't A= e^-2c?
Yes, you are correct. That is my typo. Can you continue on from there?

7. y(0)= 1/2(1-A)= 5/2
1-A = 5
A= -4

y(t) = 1/2( 1 + 4e^(-2t))

is this correct?

8. Originally Posted by linalg123
y(0)= 1/2(1-A)= 5/2
1-A = 5
A= -4

y(t) = 1/2( 1 + 4e^(-2t))

is this correct?
Looks good to me.