Complex solutions to ODEs

Question

How do I solve the following IVP problem in Mathematica so that I get real solutions?

$Q'(t)=b - \dfrac{Q(t)}{100-t}; \quad Q(0)=250$

I tried the following:

$\text{$\$$Assumptions}=b>0;\text{$\$$Assumptions}=t>0;$

$f=\text{DSolve}\left[\left\{Q'(t)=b-\frac{Q(t)}{100-t},Q(0)=250\right\},Q,t\right][[1,1,2]]$

$f(t)$

which results in the following:

$\frac{1}{2} (-2 i \pi b t-2 b t \log (100)-200 b \log (t-100)+2 b t \log (t-100)+200 i \pi b+200 b \log (100)-5 t+500)$

Any help would be much appreciated. Thanks!!!

Answer 1

You can get complex solution, depending on your initial condition !

First solve without setting IC

ode = q'[t] == b - q[t]/(100 - t);
ic = q[0] == 250;
sol = q[t] /. First@DSolve[ode, q[t], t]

The above is your solution q(t). It is all nice and no complex numbers. But you want the above to be 250 when t=0, so

myConstant =C[1]/. First@Solve[(sol/.t->0)==0,C[1]]

(sol/.C[1]->myConstant)//Simplify

So the complex solution comes from your initial conditions requirements at t=0. If you change initial conditions to something else, the solution is not complex. Push the time to over 100

ode=q'[t]==b- q[t]/(100-t);
ic=q[0]==250;
sol=q[t]/.First@DSolve[{ode,q[101]==250},q[t],t]

And now the solution is real. So it depends on where you set the initial condition at and the complex solution comes from solving for constant of integration.

Answer 2

Another way is as follows.

$Assumptions = b > 0; $Assumptions = t >= 0;
DSolve[{Q'[t] == b \[Minus] Q[t]/(100 \[Minus] t), Q[0] == 250}, Q[t], t]

{{Q[t] -> 1/2 (500 + 200 I b [Pi] - 5 t - 2 I b [Pi] t + 200 b Log[100] - 2 b t Log[100] - 200 b Log[-100 + t] + 2 b t Log[-100 + t])}}

 FullSimplify[1/2 (500 + 200 I b \[Pi] - 5 t - 2 I b \[Pi] t + 200 b Log[100] - 
2 b t Log[100] - 200 b Log[-100 + t] + 2 b t Log[-100 + t]),Assumptions->b > 0&& t> 0 && t <= 100]

-(1/2) (-100 + t) (5 + b Log[10000] - 2 b Log[100 - t])