First order linear differential equationsEdexcel International A Level Further Maths: Revision notes

A first order linear differential equation can be written in the standard form $\frac{dy}{dx}+P(x)\,y=Q(x),$ where $P$ and $Q$ are functions of $x$ (or constants). The integrating factor is $I(x)=e^{\int P\,dx}$ (no constant is needed when finding $I$; it may be quoted without proof). Multiplying the equation by $I$ makes the left side the derivative of a product: $\frac{d}{dx}\bigl(yI\bigr)=QI.$ Integrate both sides, remembering the constant $C$, and divide by $I$ to find $y$. Example: $\frac{dy}{dx}+2y=4x$. Then $I=e^{2x}$ and $\frac{d}{dx}(ye^{2x})=4xe^{2x}$. Integrating by parts gives $ye^{2x}=2xe^{2x}-e^{2x}+C$, so $y=2x-1+Ce^{-2x}$.

The coefficient of $\frac{dy}{dx}$ must be $1$. Divide every term by the coefficient first. Example: $x\frac{dy}{dx}+3y=x^2$ for $x>0$ becomes $\frac{dy}{dx}+\frac3xy=x$. Then $I=e^{\int\frac3x\,dx}=e^{3\ln x}=x^3$, so $\frac{d}{dx}(x^3y)=x^4$ and $x^3y=\frac{x^5}{5}+C$. Hence $y=\frac{x^2}{5}+\frac{C}{x^3}$. Useful identities for simplifying the integrating factor: $e^{\ln f(x)}=f(x)$ and $e^{k\ln x}=x^k$. When $x>0$ you may write $\ln x$ instead of $\ln|x|$.

The same method works when $P$ is trigonometric. Key results: $\int\tan x\,dx=\ln\sec x$ gives $I=\sec x$; $\int\cot x\,dx=\ln\sin x$ gives $I=\sin x$. Example: $\frac{dy}{dx}+y\cot x=\cos x$ for $0<x<\pi$. Then $I=\sin x$, so $\frac{d}{dx}(y\sin x)=\sin x\cos x$ and $y\sin x=\frac12\sin^2x+C$. Hence $y=\frac12\sin x+\frac{C}{\sin x}$. Use the interval given for $x$ to remove modulus signs: on $0<x<\pi$, $\sin x>0$.

Use a given condition to find $C$ after you have the general solution. Example: $\frac{dy}{dx}+2y=4x$ with $y=3$ when $x=0$. The general solution is $y=2x-1+Ce^{-2x}$, and $3=-1+C$ gives $C=4$. Hence $y=2x-1+4e^{-2x}$. Behaviour: the term $Ce^{-2x}$ (the part that decays) vanishes as $x\to\infty$, so $y\approx2x-1$ for large $x$. This long-term behaviour is the same for every value of $C$.

Many problems have the form $\frac{dm}{dt}=(\text{rate in})-(\text{rate out})$. Example: a tank holds $50$ litres of pure water. Brine with $0.2$ kg per litre enters at $5$ litres per minute, and the stirred mixture leaves at the same rate. Then rate in $=1$ kg per minute and rate out $=5\times\frac{m}{50}=\frac{m}{10}$, so $\frac{dm}{dt}+\frac{m}{10}=1$. $I=e^{t/10}$, so $me^{t/10}=10e^{t/10}+C$. With $m=0$ at $t=0$, $m=10\left(1-e^{-t/10}\right)$, which tends to $10$ kg. In evaluation, comment on the long-term value (it matches the inflow concentration times the volume) and on model assumptions such as instant uniform mixing.