First-order differential equations and integrating factorsEdexcel A-Level Further Maths: Revision notes

A first-order equation of the form $\frac{dy}{dx}+P(x)y=Q(x)$ is linear: $y$ and $\frac{dy}{dx}$ appear only to the first power. The coefficient of $\frac{dy}{dx}$ must be 1, so divide through first if necessary: $x\frac{dy}{dx}+3y=x^2$ becomes $\frac{dy}{dx}+\frac3xy=x$. If the equation can be written as $\frac{dy}{dx}=f(x)g(y)$ it is separable and you can separate the variables. When it is linear and not separable (because of a term in $y$ added to $\frac{dy}{dx}$ and a function of $x$ on the right), use an integrating factor.

The integrating factor is $I(x)=e^{\int P(x)\,dx}$ (you may quote this without proof). Multiply the equation by $I$; the left-hand side becomes $\frac{d}{dx}(Iy)$. Then integrate: $\frac{d}{dx}\left(Iy\right)=IQ\ \Rightarrow\ Iy=\int IQ\,dx.$ Example: $\frac{dy}{dx}+\frac2xy=x^2$. $\int\frac2x\,dx=2\ln x$, so $I=e^{2\ln x}=x^2$. Then $\frac{d}{dx}(x^2y)=x^4$, so $x^2y=\frac{x^5}5+c$ and $y=\frac{x^3}5+\frac c{x^2}$. Use $e^{\ln f(x)}=f(x)$ and $e^{k\ln x}=x^k$ to simplify. For $\frac{dy}{dx}+y\tan x=\cos x$, $\int\tan x\,dx=\ln\sec x$, so $I=\sec x$ and $y=(x+c)\cos x$.

The general solution contains one arbitrary constant, so it describes a family of curves. A particular solution uses a given condition, such as $y=1$ at $x=1$, to find the constant. Always divide through by the integrating factor after integrating, so the constant is divided too: $x^2y=\frac{x^5}5+c$ gives $y=\frac{x^3}5+cx^{-2}$. Sketching the family: curves from different values of $c$ never cross (the solution through each point is unique). Look at the behaviour for large $x$ (here the $\frac{x^3}5$ term dominates) and near any asymptote or point where the equation breaks down. For $\frac{dy}{dx}+y=2$, $y=2+ce^{-x}$: every member approaches the asymptote $y=2$ as $x\to\infty$, from above if $c>0$ and from below if $c<0$.

Kinematics. Acceleration is $\frac{dv}{dt}$. A parachutist with $\frac{dv}{dt}=10-0.4v$ gives $\frac{dv}{dt}+0.4v=10$, $I=e^{0.4t}$, so $v=25+ce^{-0.4t}$. With $v=0$ at $t=0$, $v=25\left(1-e^{-0.4t}\right)$. As $t\to\infty$, $v\to25$ m s$^{-1}$, the terminal speed. Distance is $\int v\,dt$. Mixing. $\frac{dm}{dt}=$ (rate in) $-$ (rate out). For a 200-litre tank with brine of 0.5 kg per litre entering at 4 litres per minute and leaving at 4 litres per minute, rate in $=2$ and rate out $=\frac{m}{200}\times4$, so $\frac{dm}{dt}+\frac m{50}=2$. Interpreting the answer: state the long-term behaviour, comment on limitations (for example, a linear model has no carrying capacity) and give units. Rates must match the units of time used.