Separable differential equationsEdexcel A-Level Maths: Revision notes

A first order differential equation is separable if it can be written $\frac{\mathrm{d}y}{\mathrm{d}x}=f(x)g(y)$. Move all $y$ terms to one side and all $x$ terms to the other, then integrate both sides: $\int\frac{1}{g(y)}\,\mathrm{d}y=\int f(x)\,\mathrm{d}x.$ Example: $\frac{\mathrm{d}y}{\mathrm{d}x}=2xy$ gives $\int\frac1y\,\mathrm{d}y=\int2x\,\mathrm{d}x$, so $\ln y=x^2+c$ and $y=Ae^{x^2}$ with $A=e^c$. Put one constant of integration, and put it in before you rearrange.

The general solution contains an arbitrary constant and represents a whole family of curves. A particular solution uses a given condition, such as $y=3$ when $x=0$, to fix the constant. For $\frac{\mathrm{d}m}{\mathrm{d}t}=-0.05m$ with $m=80$ at $t=0$: $m=Ae^{-0.05t}$ and $A=80$, so $m=80e^{-0.05t}$. Exponential decay and growth arise whenever the rate of change is proportional to the quantity: $\frac{\mathrm{d}m}{\mathrm{d}t}=km$ has solution $m=Ae^{kt}$.

Sometimes the right-hand side must be factorised before the variables can be separated: $\frac{\mathrm{d}y}{\mathrm{d}x}=xy+x=x(y+1)\ \Rightarrow\ \int\frac{1}{y+1}\,\mathrm{d}y=\int x\,\mathrm{d}x.$ This gives $\ln|y+1|=\frac{x^2}{2}+c$, so $y=Ae^{x^2/2}-1$. Look for a common factor whenever the right-hand side is a sum or difference of terms.

Varying the constant gives a family of solution curves. For $y=Ae^{x^2}$:

• every curve with $A>0$ lies above the $x$-axis, passes through $(0,A)$ and is symmetric about the $y$-axis;
• $A<0$ gives the reflections in the $x$-axis;
• $A=0$ gives the $x$-axis itself;
• the curves never cross each other, because each point $(x,y)$ fixes a single value of $A$. A sketch should show two or three members and the key feature, such as the intercept on the $y$-axis.

In context, interpret the constants and check that the solution makes sense.

• Kinematics: $a=\frac{\mathrm{d}v}{\mathrm{d}t}$, so $\frac{\mathrm{d}v}{\mathrm{d}t}=-0.4v^2$ gives $\frac1v=0.4t+\frac15$ and $v=\frac{5}{2t+1}$. Distance is $\int v\,\mathrm{d}t$.
• Limitations: this $v\to0$ as $t\to\infty$ but the distance $\frac52\ln(2t+1)$ is unbounded, so the particle never stops.
• Water draining: $\frac{\mathrm{d}h}{\mathrm{d}t}=-0.1\sqrt{h}$ gives $h=(2-0.05t)^2$, valid only until $t=40$, when the tank is empty. Always state the domain of validity: a solution may give negative or unbounded values outside the range where the model works.