## xartaan Group Title 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). one year ago one year ago

1. xartaan Group Title

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 Group Title

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 Group Title

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 Group Title

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

5. hartnn Group Title

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

6. hartnn Group Title

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

7. hartnn Group Title

i put e^c = -225

8. xartaan Group Title

ah i see

9. hartnn Group Title

any more doubts ? all clear ?

10. xartaan Group Title

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

11. xartaan Group Title

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 Group Title

welcome :)