## xartaan 2 years ago Another differential equation $\frac{ dy }{ dt }=0.4(y-300)$ and y(0)=75. So, I separate the variables into dy/(y-300 = .4dt integrate both side for ln(y-300)=.4t+c exponentiate then add 300 to the right for y=300+e^(.4t+c).

1. xartaan

Now I think I should use my initial value, so 75=300+e^c => -225=e^c => 225=-e^c , and take logs of both sides to get ln(225)=-c

2. xartaan

Next I should plug -ln(225) for c in the equation ln(y-300)=.4t+c, giving me ln(y-300)=.4t-ln(225) right? Or and I doing something wrong?

3. xartaan

I added the log from the right to the left, and using property of logs combined them to a quotient, but the answer after simplification was wrong, so I must be missing something.

4. hartnn

225=-e^c , and take logs of both sides to get ln(225)=-c <----NO

5. hartnn

if 225 = e^(-c) then -c = ln 225

6. hartnn

y=300+e^(.4t+c). y=300+e^(.4t)e^c. y=300-225e^(.4t)

7. hartnn

i put e^c = -225

8. xartaan

ah i see

9. hartnn

any more doubts ? all clear ?

10. xartaan

I think so, I am not so quick, I will work it out!

11. xartaan

Ah, there really isn't any working out to do because e^c is a constant and its all in terms of y. So simple, thanks again.

12. hartnn

welcome :)