Differential Problem with solution curve: dx/dt = r − kx, x(0) = 0

[carpenterhelper]

A mathematical model for the rate at which a substance enters the bloodstream, before the liver and kidneys start to eliminate it, is dx/dt = r − kx, x(0) = 0 where r and k are empirically derived constants. Solve the o.d.e. and sketch the graph of a typical solution curve. Be sure to clearly identify the x value of any asymptote which may be present in the solution. Finally, at what time t does the concentration reach one-half of the asymptotic maximum value?


Attempted Solution:

dx = (r-kx)dt
integral(1)dx = integral(r-kx)dt
x = rt -kxt + c

Not really sure if there is an asymptote

 

[Deleted member 4993]

A mathematical model for the rate at which a substance enters the bloodstream, before the liver and kidneys start to eliminate it, is dx/dt = r − kx, x(0) = 0 where r and k are empirically derived constants. Solve the o.d.e. and sketch the graph of a typical solution curve. Be sure to clearly identify the x value of any asymptote which may be present in the solution. Finally, at what time t does the concentration reach one-half of the asymptotic maximum value?


Attempted Solution:

dx = (r-kx)dt
integral(1)dx = integral(r-kx)dt ......... That is incorrect
x = rt -kxt + c

Not really sure if there is an asymptote

dx = (r-kx)dt

dx/(r-kx) = dt

After integrating:

ln|(r-kx)| = -kt + C

Now continue......

 

[HallsofIvy]

A mathematical model for the rate at which a substance enters the bloodstream, before the liver and kidneys start to eliminate it, is dx/dt = r − kx, x(0) = 0 where r and k are empirically derived constants. Solve the o.d.e. and sketch the graph of a typical solution curve. Be sure to clearly identify the x value of any asymptote which may be present in the solution. Finally, at what time t does the concentration reach one-half of the asymptotic maximum value?


Attempted Solution:

dx = (r-kx)dt
integral(1)dx = integral(r-kx)dt
x = rt -kxt + c

You can't integrate like that because x is not a constant!

Not really sure if there is an asymptote