# Solution to differential equation

I apologise for the lack of LaTeX but I have no clue how to use it

Anyway - I've been searching for ages on how to solve this problem none of my textbooks seem to give any solutions

How would I solve

$$\frac{dv}{dt} + Bv^2 = -A$$

If someone could also give me what type of differential equation this is it would be much appreciated

Thanks

• if $AB<0$ : $v(t)=\sqrt{\dfrac{-A}{B}}+\dfrac{1}{Bt+c}$ appears to do the trick for all values of $c$ – imj Jan 4 '14 at 20:10
• "I've been searching for ages on how to solve this problem none of my textbooks seem to give any solutions" Then change your textbooks. – Did Jan 4 '14 at 20:24
• There is a serious notational issue. $B(v^2)$ looks like an arbitrary function of $v^2$. We're all assuming this is just a constant times $v^2$. You need to sort such things out in your head, for starters. With an arbitrary function there, you can't possibly solve the equation explicitly. – Ted Shifrin Jan 4 '14 at 20:26

## 2 Answers

Assuming $A$ and $B$ are constants, just do the usual separation of variables computation (here I assume $A,B>0$, but if $A,B<0$, use the negative): \begin{align*} \frac{dv}{dt} &= -(A+Bv^2) \\ \frac{dv}{A+Bv^2} &= -dt \\ \int \frac{dv}{A+Bv^2} &= -\int dt \,, \end{align*} and, indeed, integrate the left hand side by making a trigonometric substitution $\sqrt Bv = \sqrt A\tan u$.

If $A$ and $B$ have opposite signs (say $A>0$, $B<0$), then you can factor $A+Bv^2 = (\sqrt A + \sqrt{-B}v)(\sqrt A - \sqrt{-B}v)$ and use partial fractions.

Assuming that $B(v^2)$ means $B\cdot v^2$ and that $A$ and $B$ are both positive, the change of variable $$v(t)=\sqrt{\frac{A}B}\cdot\tan u(t)$$ yields $$u'(t)=-\sqrt{AB},$$ hence, for every $t$, $$v(t)=\sqrt{\frac{A}B}\cdot\tan\left(\arctan\left(v(0)\cdot\sqrt{\frac{B}A}\right)-\sqrt{AB}\cdot t\right).$$ If $A$ and $B$ are negative, changing $t$ into $-t$ in $u(t)$ above yields the solution. If $AB\lt0$, using $\tanh$ instead of $\tan$ yields the solution.