Where do we need absolute values when solving $u'(t) = \frac{u^2(t) - u(t)}{t}$

Question

Solve the differential equation $$ u'(t) = \frac{u^2(t) - u(t)}{t} $$ and then solve the initial value problem with $u(1) = \frac{1}{2}$.

I know the general solution is given by $u(t) = \frac{1}{c_1 t + 1}$ for all $t \in \mathbb{R}$ and some $c_1 \in \mathbb{R}$ determined by the initial condition, which is not always positive.

My problem is that when I integrate $\frac{1}{t}$ when using separation of variables, is my result $\ln|t|$ or just $\ln(t)$?

Here's what I've done so far: We exclude the constant solutions $u :\equiv 0$ and $u : \equiv 1$ Then our differential equation is equivalent to $$ \frac{u’(t)}{u^2(t) - u(t)} = \frac{1}{t}. $$ Integrating and substituting gives \begin{align*} \int_{u(t_0)}^{u(t)} \frac{1}{s(s - 1)} ds = \int_{t_0}^{t} \frac{1}{s} ds \implies & \int_{u(t_0)}^{u(t)} \frac{1}{s - 1} - \frac{1}{s} ds = \ln\left| \frac{t}{t_0} \right| \\ \implies & \ln\left|\frac{u(t_0) ( u(t) - 1 )}{u(t) ( u(t_0) - 1)} \right| = \ln\left| \frac{t}{t_0} \right| \\ \implies & \frac{| u(t) |}{|u(t) - 1|} = \frac{|t_0| |u(t_0)|}{|t| | u(t_0) - 1|} \end{align*} Do I now do a case distinction or how can I solve for $u$ now?

Answer 1

The absolute values in your calculation were applied properly where they are needed. However, in the next step you can remove them as follows:

As you observed, there are constant solutions at $u=0$ and $u=1$. Due to uniqueness of solutions any solution starting in $(0,1)$ stays inside that interval, the same for $(-\infty,0)$ and $(1,\infty)$. Thus the signs of the expressions under the absolute values are constant and can be merged into the integration constant. Resp. in a formulation with the initial condition, those fractions $\frac{u(t)}{u(t_0)}$ and $\frac{u(t)-1}{u(t_0)-1}$ are always positive, so that the absolute value is equal to the term within.

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The ODE has a singularity at $t=0$, thus the whole vertical line there is excluded from the domain of the ODE. Thus the domain of the solution with initial condition at $t=1$ is $(0,\infty)$, not the whole of $\Bbb R$. So also $\frac{t}{t_0}$ is positive over the domain of the solution.

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Then you continue with (with the correct signs in the partial fraction decomposition) \begin{align} \frac{u(t_0)(u(t)-1)}{u(t)(u(t_0)-1)}&=\frac{t}{t_0}\\[.8em] u(t)\Bigl(t_0u(t_0)-t(u(t_0)-1)\Bigr)&=t_0u(t_0)\\[.8em] u(t)=\frac{t_0u(t_0)}{t_0u(t_0)+(1-u(t_0))t} \end{align}

Answer 2

Hint.

$$ (t u)' = u^2 $$

now making $ v = t u$

$$ v' = \frac{v^2}{t^2}\Rightarrow \frac{dv}{v^2} = \frac{dt}{t^2} $$

or integrating

$$ \frac 1v = \frac 1t + C\Rightarrow v = \frac{t}{C t+1} = u t $$

Answer 3

The maximal $(t,u)$-domain relevant to the given IVP is $\Omega:={\mathbb R}_{>0}\times{\mathbb R}$. Within $\Omega$ the standard existence and uniqueness theorem for ODEs is valid. By inspection one sees that there are the constant solutions $u_0(t)\equiv0$ and $u_1(t)\equiv1$ $(t>0)$. No other solution can cross the graphs of $u_0$ and $u_1$. Since the initial point $\bigl(1,{1\over2}\bigr)$ is given the solution $t\mapsto u(t)$ therefore stays in the $u$-interval $\>]0,1[\>$ for all $t>0$. Hence there are no case distinctions; you just have to select the primitive of $s\mapsto{1\over s^2-s}$ that is valid for $0<s<1$, namely $$\eqalign{\int{ds\over s^2-s}&=\int\left({1\over s-1}-{1\over s}\right)ds=\log|s-1|-\log |s| +C\cr &=\log{1-s\over s}+C\qquad(0<s<1)\ .\cr}$$