# Separable equations

## Solving separable equations

• A differential equation is a separable equation, if it can be written in the form $\frac{\mathrm dy}{\mathrm dx}=f(x)g(y).$
• In some cases, it might be easier to recognize it as written in the form $\frac{\mathrm dy}{\mathrm dx}=\frac{f(x)}{h(y)}$
• To solve a separable equation, we transform its equation so that each side of the equation only has one variable (hence the name separable): $\frac{\mathrm dy}{g(y)}=f(x)\,\mathrm dx,$
• and then we integrate: $\int\frac{\mathrm dy}{g(y)}=\int f(x)\,\mathrm dx.$
• This gives an implicit equation for the solution $$y=y(x)$$. In lucky cases, we can solve this equation for $$y$$.
• Example. consider the logistic DE: $\frac{\mathrm dP}{\mathrm dt}=kP\left(1-\frac{P}{M}\right).$
• This is an autonomous equation, since $$t$$ only appears as the variable of $$P$$ in it. In particular, it is separable.
• We can separate variables: $\frac{\mathrm dP}{P(1-P/M)}=k\mathrm dt$
• and integrate: $\int\frac{\mathrm dP}{P(1-P/M)}=\int\frac{M}{P(M-P)}\,\mathrm dP=\int\frac{1}{P}+\frac{1}{M-P}\,\mathrm dP=\ln|P|-\ln|M-P|+C=\ln\left|\frac{P}{M-P}\right|+C.$
• We have gotten the implicit equation which in this case we can solve: \begin{align*} \ln\left|\frac{P}{M-P}\right|+C&=kt\\ \ln\left|\frac{M-P}{P}\right|&=-kt+C\\ \left|\frac{M-P}{P}\right|&=e^Ce^{-kt}\tag{A=e^C}\\ \frac{M-P}{P}&=Ae^{-kt}\\ \frac{M}{P}-1&=Ae^{-kt}\\ P&=\frac{1}{1+Ae^{-kt}}. \end{align*}
• Exercises. 9.3: 1, 5, 15, 19

## Orthogonal trajectories

• Recall that if line $$L_1$$ has slope $$m$$: $L_1:(y-y_1)=m(x-x_1),$ and line $$L_2$$ is perpendicular to $$L_1$$, then $$L_2$$ has slope $$-m^{-1}$$: L_2:(y-y_2)=-m^{-1}(x-x_2).
• In orthogonal trajectories, given a family of curves, we can set up a differential equation the solutions of which are curves perpendicular to every member of the original family.
• For example, consider the family of parabolas $$x=ky^2$$ where $$k$$ is a constant.
• To get a differential equation the solutions of which are exactly these curves, we implicitly derivate by $$x$$: $1=2kyy'\text{ or }y'=\frac{1}{2ky}.$
• Since $$x=ky^2$$ yields $$k=\frac{x}{y^2}$$, we can eliminate $$k$$: $y'=\frac{y}{2x}.$
• Now to get orthogonal curves, we need to change the slopes to the perpendicular ones: $y'=-\frac{2x}{y}.$
• Separate variables: $y\,\mathrm dy=-2x\,\mathrm dx$
• Integrate: $\frac{y^2}{2}=-x^2+C.$
• We see that the orthogonal trajectories are the ellipses $x^2+\frac{y^2}{2}=C.$
• Exercises. 9.3: 29, 30.

## Mixing problems

• A common problem type in separable equations is that of mixing problems.
• It is about a tank containing a thoroughly mixed solution of some substance, for example salt.
• You are told how a solution of a possibly different density is entering the tank, and how the thoroughly mixed solution is exiting the tank.
• Letting $$y(t)$$ denote the amount of the substance in the tank at time $$t$$, using the problem description, you need to create a differential equation in the form $\frac{\mathrm dy}{\mathrm dt}=\text{(rate in)}-\text{(rate out)}$ and solve it.
• Example. A tank contains 20 kg of salt dissolved in 5000 L of water. Brine that contains 0.03 kg of salt per litre of water enters the tank at a rate of 25 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt remains in the tank after half an hour?
• Rate in: 25 L/min of 0.03 kg/L salt is incoming, so we get $$25\cdot 0.03=\frac34$$ kg/min.
• Rate out: the tank contains $$y(t)$$ kg salt. Since exit rate is 25 L/min, we get $$y(t)\cdot\frac{25}{5000}$$ kg/min.
• Our differential equation is $\frac{\mathrm dy}{\mathrm dt}=\text{(rate in)}-\text{(rate out)}=\frac34-\frac{y}{200}=\frac{150-y}{200}.$
• Separate variables and integrate: \begin{align*} \int\frac{\mathrm dy}{150-y}&=\int\frac{\mathrm dt}{200}\\ -\ln|150-y|&=\frac{t}{200}+C \end{align*}
• Since $$y(t=0)=20$$ gives $$-\ln130=C$$, we get \begin{align*} -\ln|150-y|&=\frac{t}{200}-\ln130\\ |150-y|&=130e^{-t/200}\\ \end{align*}
• Since the RHS is always nonnegative, so is the LHS, therefore we get \begin{align*} y(t)&=150-130e^{-t/200}. \end{align*}
• The amount of salt after half an hour is $y(t=30)=150-130e^{-30/200}.$
• Exercises. 9.3: 45, 46, 48 (I know you like odd-numbered exercises, but 47 is utterly disgusting)