Separable first order differential equationsEdexcel International A Level Maths: Revision notes

A first order differential equation is separable if it can be written $\frac{dy}{dx}=f(x)g(y)$: the right-hand side is a product of a function of $x$ only and a function of $y$ only. Then you move all the $y$ terms to one side with $dy$ and all the $x$ terms to the other with $dx$: $\int\frac{1}{g(y)}\,dy=\int f(x)\,dx.$ Example: $\frac{dy}{dx}=\frac{x}{y}$ becomes $\int y\,dy=\int x\,dx$, so $\frac{y^2}{2}=\frac{x^2}{2}+c$, i.e. $y^2=x^2+A$.

Integrating gives a general solution, which contains one arbitrary constant and describes a whole family of curves. Put a single $+c$ on one side only; two constants merge into one. For $\frac{dy}{dx}=2xy$: $\int\frac1y\,dy=\int2x\,dx$ gives $\ln|y|=x^2+c$, so $y=e^{x^2+c}=Ae^{x^2}$ where $A=\pm e^c$. Useful integrals: $\int\frac{1}{ay+b}\,dy=\frac1a\ln|ay+b|$, $\int y^n\,dy=\frac{y^{n+1}}{n+1}$, $\int e^{ay}\,dy=\frac1ae^{ay}$. For $\frac{dy}{dx}=e^{x-y}$, write $e^{x-y}=e^xe^{-y}$ so $\int e^y\,dy=\int e^x\,dx$, giving $e^y=e^x+c$.

A particular solution is the member of the family that passes through a given point, called an initial condition or boundary condition. Substitute the values to find $c$, then rearrange. Example: $\frac{dy}{dx}=\frac{2y+1}{x}$, $x>0$, passes through $(1,2)$. $\int\frac1{2y+1}\,dy=\int\frac1x\,dx$ gives $\frac12\ln|2y+1|=\ln x+c$. At $(1,2)$: $\frac12\ln5=c$. So $\ln(2y+1)=2\ln x+\ln5=\ln(5x^2)$, hence $2y+1=5x^2$ and $y=\frac{5x^2-1}{2}$. Write constants as logarithms when it helps combine terms, then use the log laws before exponentiating.

Rates of change in context often lead to separable equations. In exponential change, $\frac{dN}{dt}=kN$ gives $N=Ae^{kt}$; $k>0$ is growth and $k<0$ decay. In Newton's law of cooling, $\frac{d\theta}{dt}=-k(\theta-\theta_0)$ gives $\theta=\theta_0+Ae^{-kt}$, with $\theta\to\theta_0$ as $t\to\infty$. For a draining tank, $\frac{dh}{dt}=-k\sqrt h$ gives $\int h^{-1/2}\,dh=\int-k\,dt$, so $2\sqrt h=-kt+c$. With $h=4$ at $t=0$ and $h=1$ at $t=10$: $c=4$, $k=0.2$, and the tank empties when $2\sqrt h=0$, at $t=20$. Use the data in order: first the initial condition for $c$ (or $A$), then the second condition for $k$.