Separable differential equations and families of solutionsEdexcel International A Level Further Maths: Revision notes

A first order equation is separable if it can be written $\frac{dy}{dx}=f(x)g(y)$. Then $\int\frac{1}{g(y)}\,dy=\int f(x)\,dx.$ Integrate each side and include one arbitrary constant. The result is the general solution. Example: $\frac{dy}{dx}=2xy$ with $y>0$. Then $\int\frac1y\,dy=\int2x\,dx$, so $\ln y=x^2+c$ and $y=Ae^{x^2}$, where $A=e^c>0$ (or any non-zero constant, allowing $y<0$). The constant of integration usually goes inside the logarithm or exponent, so it becomes a multiplier $A$ after exponentiating. Give the answer as $y=\ldots$ when asked, but an implicit form is acceptable when $y$ cannot be isolated easily. If $g(y)=0$ for some value, then $y=$ that constant is also a solution and may be lost in the division.

A particular solution is the member of the family through a given point. Substitute the given $x$ and $y$ into the general solution to find the constant. Example: $\frac{dy}{dx}=xe^{-y}$ with $y=0$ when $x=0$. Multiplying by $e^y$ gives $e^y\frac{dy}{dx}=x$, so $\int e^y\,dy=\int x\,dx$ and $e^y=\frac{x^2}{2}+C$. At $(0,0)$, $1=C$, so $y=\ln\left(1+\frac{x^2}{2}\right)$. After finding $y$, state the domain on which it is valid (for example $x>0$ when $\ln x$ appears, or $x\neq e$ when a denominator vanishes), and use the equation to find gradients and stationary points directly without differentiating the solution.

To form a differential equation from a description:

1. Name the variables and write the rate as $\frac{dy}{dt}$ (or $\frac{dy}{dx}$ for a gradient).
2. "is proportional to" means "$=k\times$"; "rate of decrease" means a negative sign.
3. Say what $k$ is and its sign. Example: a drink cools in a room at $20\,^\circ\text{C}$ and its rate of decrease is proportional to $\theta-20$. Then $\frac{d\theta}{dt}=-k(\theta-20)$ with $k>0$. Separating gives $\ln(\theta-20)=-kt+c$, so $\theta=20+Ae^{-kt}$. If $\theta=80$ at $t=0$ then $A=60$. For a geometric condition such as "the gradient at each point equals the $y$-coordinate divided by $x^2$", write $\frac{dy}{dx}$ as the stated expression. Use the particular point given to find the constant.

The arbitrary constant gives a family of curves, one for each value of the constant. Through any point where the differential equation is defined there is exactly one member of the family, so members never cross.

• $\frac{dy}{dx}=2xy$: $y=Ae^{x^2}$. All curves cross the $y$-axis at $(0,A)$ with a minimum there when $A>0$ and a maximum when $A<0$; $A=0$ gives $y=0$.
• $\frac{dy}{dx}=-\frac xy$: $y\,dy=-x\,dx$ gives $x^2+y^2=k$, a family of circles centred at the origin.
• $x\frac{dy}{dx}=y^2$: $y=\frac{1}{c-\ln x}$, with a vertical asymptote at $x=e^c$ and the asymptote $y=0$ as $x\to\infty$. To sketch members, identify intercepts and asymptotes, use the sign of $\frac{dy}{dx}$ to find where curves rise or fall, locate stationary points where $\frac{dy}{dx}=0$, and draw two or three members for different values of the constant.

Always check a solution: differentiate your $y$ and substitute into the original equation, or test the initial condition. Use $\ln|y|$ when $y$ may be negative; if $y>0$ is given, write $\ln y$. State the final answer in the form requested ($y=\ldots$ or $e^y=\ldots$). In modelling questions, interpret the constants: $k$ is a rate constant with units (for example per minute), and the model may break down for large $t$. Standard integrals needed: $\int\frac1y\,dy=\ln|y|$, $\int e^y\,dy=e^y$, $\int y^{-2}\,dy=-\frac1y$.