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

A first-order linear differential equation can be written in the standard form $\frac{dy}{dx}+P(x)y=Q(x).$ The coefficient of $\frac{dy}{dx}$ must be 1, so divide through first if necessary: $x\frac{dy}{dx}+3y=x^3$ becomes $\frac{dy}{dx}+\frac3xy=x^2$. The method is appropriate when $y$ and $\frac{dy}{dx}$ each appear only to the power 1 and the equation cannot be separated into $x$ and $y$ terms. Terms like $y^2$ or $y\frac{dy}{dx}$ rule it out.

The integrating factor is $I(x)=e^{\int P(x)\,dx}.$ Multiplying the equation by $I$ makes the left-hand side an exact derivative: $\frac{d}{dx}\left(I\,y\right)=I\,Q.$ Steps: (1) write in standard form; (2) find $P$ and then $I$, with no constant of integration in $I$; (3) multiply every term by $I$; (4) write the left side as $\frac{d}{dx}(Iy)$; (5) integrate both sides, adding $+c$; (6) divide by $I$. Simplify $e^{\ln f}=f$ and $e^{-\ln f}=\frac1f$ when forming $I$.

Solve $\frac{dy}{dx}+\frac3xy=x^2$ for $x>0$. $P=\frac3x$, so $I=e^{\int\frac3x\,dx}=e^{3\ln x}=x^3$. Multiplying: $x^3\frac{dy}{dx}+3x^2y=x^5$, i.e. $\frac{d}{dx}(x^3y)=x^5$. Integrating: $x^3y=\frac{x^6}{6}+c$. So $y=\frac{x^3}{6}+\frac{c}{x^3}.$ A second example, $\frac{dy}{dx}+y\tan x=\cos x$, has $I=e^{\ln\sec x}=\sec x$, so $\frac{d}{dx}(y\sec x)=1$ and $y=(x+c)\cos x$.

The general solution contains one arbitrary constant, because one integration is performed. A particular solution is found by substituting a given condition, such as $y=3$ when $x=0$, into the general solution to find $c$. Always find the general solution first, then use the condition, then write $y=f(x)$. For $y=(x+c)\cos x$ with $y=3$ at $x=0$: $3=c$, so $y=(x+3)\cos x$.

In kinematics, $a=\frac{dv}{dt}$, so equations such as $\frac{dv}{dt}+2v=10e^{-t}$ model motion with a resistance proportional to velocity. Here $I=e^{2t}$, giving $\frac{d}{dt}(ve^{2t})=10e^{t}$, so $v=10e^{-t}+ce^{-2t}$. If $v=0$ at $t=0$, $c=-10$ and $v=10e^{-t}-10e^{-2t}$. Maximum velocity occurs when $\frac{dv}{dt}=0$, at $t=\ln2$ with $v=2.5$. Integrate $v$ once more for displacement, using $s=0$ at $t=0$ to fix the constant. Interpret limits: as $t\to\infty$, $e^{-t}\to0$, so the particle approaches a fixed position.