5.14 Differential equations: separation of variablesIB Maths: Applications and Interpretation HL: Revision notes

A differential equation links a quantity to its rate of change. The words proportional to give a constant of proportionality $k$:

• growth of $G$ at a rate proportional to $G$: $\frac{dG}{dt}=kG$ with $k>0$.
• decay of $m$ at a rate proportional to $m$: $\frac{dm}{dt}=-km$ with $k>0$.
• cooling towards a room temperature of $20$ °C: $\frac{dT}{dt}=-k(T-20)$. Define every symbol with units. The minus sign shows decrease, so keep $k$ positive.

If $\frac{dy}{dx}=f(x)g(y)$, separate so that all the $y$ terms are on one side and all the $x$ terms on the other, then integrate: $\int\frac{1}{g(y)}\,dy=\int f(x)\,dx.$ Steps: (1) separate, (2) integrate both sides, with one constant $c$, (3) rearrange for $y$ if asked. Example: $\frac{dy}{dx}=ky\Rightarrow\int\frac1y\,dy=\int k\,dx\Rightarrow\ln y=kx+c\Rightarrow y=Ae^{kx}$, where $A=e^c$.

The general solution contains an arbitrary constant, e.g. $G=Ae^{kt}$; it describes a whole family of curves. A particular solution uses an initial condition to find the constant, e.g. $G=50$ when $t=0$ gives $A=50$, so $G=50e^{kt}$. Use a second piece of information to find $k$. If $G=80$ when $t=4$, then $50e^{4k}=80$ and $k=\frac{\ln1.6}{4}=0.118$. The exponential model $y=Ae^{kx}$ is the solution of $\frac{dy}{dx}=ky$: it grows if $k>0$ and decays if $k<0$.

A sample has mass $m$ mg with $\frac{dm}{dt}=-km$, $m=200$ when $t=0$ and $m=150$ when $t=5$. Separate: $\int\frac1m\,dm=-k\int dt$, so $\ln m=-kt+c$ and $m=Ae^{-kt}$. Initially $A=200$. Then $150=200e^{-5k}$, so $e^{-5k}=\frac34$ and $k=\frac{\ln(4/3)}{5}=0.0575$. The mass after $20$ years is $200e^{-20k}=200\left(\frac34\right)^4=63.3$ mg.

$\frac{dv}{dt}=-0.05v^2$ with $v=10$ when $t=0$. Separate: $\int v^{-2}\,dv=\int-0.05\,dt$, so $-\frac1v=-0.05t+c$. When $t=0$, $-\frac{1}{10}=c$, so $\frac1v=0.05t+0.1$ and $v=\frac{20}{t+2}$. For Newton's law of cooling, $\frac{dT}{dt}=-k(T-20)$: $\int\frac{1}{T-20}\,dT=-k\int dt$, so $T=20+Ae^{-kt}$. With $T=90$ at $t=0$, $A=70$. In the long term $e^{-kt}\to0$, so $T\to20$.

Always link the maths back to the context. In $G=50e^{kt}$, $50$ g is the initial mass and $k$ is the growth rate parameter (per day). State units and give 3 s.f. Say when a model has limits: pure exponential growth predicts unlimited growth, which is unrealistic for a real pond. To find the time to reach a target value, set the solution equal to it and take natural logarithms. Doubling time of $\frac{dG}{dt}=kG$ is $\frac{\ln2}{k}$.