Separable differential equationsAQA A-Level Maths: Revision notes

A first-order differential equation is separable if it can be written as $\frac{dy}{dx}=f(x)\,g(y)$. Divide by $g(y)$, treat $dy$ and $dx$ as separate, and integrate each side: $\int\frac{1}{g(y)}\,dy=\int f(x)\,dx.$ Example: $\frac{dy}{dx}=2xy$ with $y>0$. Then $\int\frac1y\,dy=\int2x\,dx$, so $\ln y=x^2+c$. Always include one constant of integration, on either side.

The solution with an arbitrary constant is the general solution; substituting a given condition (such as $y=3$ when $x=0$) fixes the constant and gives the particular solution. From $\ln y=x^2+c$: $y=e^{x^2+c}=Ae^{x^2}$, where $A=e^c$. This step is the usual place to deal with the constant: write the constant before you exponentiate. If $y=3$ when $x=0$ then $A=3$, so $y=3e^{x^2}$. When you integrate $\frac{1}{y}$ you get $\ln|y|$; if the context says $y>0$ you may drop the modulus.

Sometimes the right-hand side must be factorised first. For $\frac{dy}{dx}=6x+3xy$, take out the common factor: $\frac{dy}{dx}=3x(2+y)$. Then $\int\frac{1}{2+y}\,dy=\int3x\,dx$, giving $\ln|2+y|=\frac32x^2+c$ and $2+y=Ae^{\frac32x^2}$. With $y=1$ when $x=0$: $1+2=A$, so $y=3e^{\frac32x^2}-2$. To find $x$ when $y=10$: $e^{\frac32x^2}=4$, so $x^2=\frac23\ln4$ and $x=0.961$.

Many models give separable equations. Rates of change such as $\frac{dP}{dt}=kP$ (growth or decay) and Newton's law of cooling $\frac{d\theta}{dt}=-k(\theta-\theta_0)$ both separate. In kinematics, acceleration $a=\frac{dv}{dt}$ and velocity $v=\frac{ds}{dt}$, so a resistive force proportional to velocity gives $\frac{dv}{dt}=-kv$ and $v=v_0e^{-kt}$. Worked example: $\frac{d\theta}{dt}=-k(\theta-20)$, $\theta=90$ at $t=0$. Then $\theta-20=70e^{-kt}$. If $\theta=60$ when $t=5$, $e^{-5k}=\frac47$ and $k=\frac15\ln\frac74=0.112$.

After solving, interpret your answer in the context: say what the constants mean (e.g. $A$ is the initial velocity) and what happens as $t\to\infty$. For $v=8e^{-0.5t}$, $v>0$ for all $t$ and $v\to0$, so the model never lets the particle stop. State limitations: the model may ignore other forces (such as constant friction), assume a constant room temperature, or only be valid over a restricted domain (for example, a population cannot grow exponentially for ever). Also check the answer is sensible, such as rejecting negative values of $x$ or $t$ where the domain forbids them.