Solving the two ODEs $\frac {dv}{dt}+\frac va=F$ and $c-\frac 12 \frac {du}{dt}-\frac ua=0$

Question

I am reading the article "Optimal Velocity in a Race" by J.B. Keller. In the article the author solves two ODEs. The equations are:
$$(1) \qquad \qquad \frac {dv}{dt}+\frac va=F \quad \text {where} \quad v(0)=0 \quad \text {and} \quad a,F>0$$ $$(2) \qquad \qquad c-\frac 12 \frac {du}{dt}-\frac ua=0 \quad \text {where} \quad t\in [t_1,t_2] \quad \text {and} \quad c,a>0$$

The solution for $(1)$ is $$v(t)=Fa(1-\exp(-\frac ta))$$ and the solution for $(2)$ is $$u(t)=ac+(u(t_1)-ac)\exp(\frac {-2t}a)$$

It was quite a long time ago since I last solved ODEs. Can someone please give some explanations on how the author solved these two eqautions? There are no details in the article.

Answer 1

Yes thats correct for the first equation $$\frac {dv}{dt}+\frac va=F$$ $$\frac {dv}{dt}=-\frac va+F$$ $$\frac {dv}{dt}=-\frac {(v-aF)}a$$ $$\int \frac {dv}{v-aF}=-\int \frac {dt}a$$ $$\ln|{v-aF}|=-\frac {t}a+K$$ $$v=Ke^{-\frac {t}a}+aF$$ $$v(0)=0 \implies K=-aF$$ $$\boxed {v=aF(1-e^{-\frac {t}a})}$$ Same method for the seond equation... $$c-\frac 12 \frac {du}{dt}-\frac ua=0$$ $$ \frac {du}{dt}=-2(\frac ua-c)$$ $$ \int \frac {du}{u-ac}=-2\int \frac {dt}a$$ $$ \ln|{u-ac}|=-2 \frac {t}a+K$$ $$ u=Ke^{-2 \frac {t}a}+ac$$ Normally the constant K should be equal to $$u=Ke^{-2 \frac {t}a}+ac \implies K=(u(t_1)-ac)e^{2 \frac {t_1}a}$$ But the author has for some reason chosen $$ K=(u(t_1)-ac)$$

$$.....$$

Answer 2

The first one is $$\dot{v}+\frac{1}{a}v=F$$ We find integrating factor which is $\exp\left(\displaystyle \int\dfrac{1}{a}\text{d}t\right)=\exp\left(\dfrac{1}{a}t\right)$, i.e we can write $$\exp\left(\frac{1}{a}t\right)\dot{v}+\exp\left(\frac{1}{a}t\right)\frac{1}{a}v=F\exp\left(\frac{1}{a}t\right)\iff \text{d}\left(\exp\left(\frac{1}{a}t\right)v\right)=F\exp\left(\frac{1}{a}t\right)$$ Integrating both sides we get $$\exp\left(\frac{1}{a}t\right)v=aF\exp\left(\frac{1}{a}t\right)+C\iff v=aF+C\exp\left(-\frac{1}{a}t\right)$$ Using initial condition $v(0)=0$, we find that $C=-aF$ and all in all $$v(t)=aF\left(1-\exp\left(-\frac{t}{a}\right)\right)$$

Second equation can be solved using same method.

Edit: as suggested in comments, the equation is separable and that suggests simpler solution.