Diffrential equation subject qs

[ishterz]

Hello,

I managed to solve the differential equation : dx/dt= 0.1 (x-250)

with the information when t=0 x=1000 and dx/dt= 75, I also found "C" and got
ln lx-250l = 0.1t + ln750

However, I am having trouble obtaining the expression for x in terms of t

I got x= e^0.1t +1000 which is wrong

Please help

Thank you for your time

 

[epenguin]

In

ln lx-250l = 0.1t + ln750

nothing authorised the ln750 - maybe you were trying to take too many steps at a time. Just write a general integration constant K, or better lnK and then find out what it must be and it all works out.

 

[HallsofIvy]

You get $ln|x-250|= 0.1t+ C$. Setting t= 0 and x= 1000, you get $ln(1000- 250)= ln(750)= C$ so, contrary to epenguin, you are correct- $ln|x- 250|= 0.1t+ 250$.

Now, take the exponential of both sides,
$$x- 250= e^{0.1t+ ln(750)}= e^{0.1t}e^{ln(750)}= 750 e^{0.1t}$$
so that x= 750 e^{0.1t}+ 250.

I suspect you made the mistake of thinking that $e^{a+ b}= e^a+ e^b$ rather than the correct $e^{a+b}= e^ae^b$.

It does happen that that $dx/dt(0)= 75[\itex] but since the equation is first order, integrating gives one undetermined constant so you can only impose one condition, not two.$