Separable differential equations (HL)

Separable differential equations in the form

\frac{\d y}{\d x} = \frac{f(x)}{g(y)}

are solved via

\int f(x)\d x = \int g(y)\d y

when a point $(x_0, y_0)$ on the solution function is given.

This can equivalently be written as a definite integral

\int_{x_0}^x f(w)\d w = \int_{y_0}^y g(\tau)\d \tau

though it may be less intuitive to work with.

Example: (May 2009 HL TZ1 Paper 2 #8) Solve the differential equation

\frac{\cos^2 x}{\e^y} - \e^{\e^y}\frac{\d y}{\d x} = 0

given that $y = 0$ when $x = \pi$ .

\begin{align*}\frac{\cos^2 x}{\e^y} - \e^{\e^y}\frac{\d y}{\d x} &= 0 \\ \frac{\cos^2 x}{\e^y} &= \e^{\e^y}\frac{\d y}{\d x} \\ \frac{\cos^2 x}{\e^y\cdot \e^{\e^y}} &= \frac{\d y}{\d x} \\ \int \cos^2 x\d x &= \int \e^y \cdot \e^{\e^y}\d y \end{align*}

The left side requires $\cos2x = 2\cos^2x-1$ and right side is in the form $f^\prime(y)\e^{f(y)}$ . It may be a good time to say that $a^{b^c} = a^{\left(b^c\right)} \neq \left(a^b\right)^c$

\begin{align*}\text{LHS} &= \int \left(\frac12 \cos 2x + \frac12\right)\d x \\ &= \frac14 \sin2x + \frac12x + C \end{align*}

\begin{align*}\text{RHS} &= \int \left(\frac{\d }{\d y}\e^y \cdot \e^{\e^y}\right)\d y \\ &= \e^{\e^y} + D \end{align*}

The general solution is

\frac14 \sin2x + \frac12x = \e^{\e^y} + E

where $C, D, E$ are constants, you could also use $C_1, C_2, C_3$ but it gets messy under time pressure. It’s entirely acceptable to make up new constants, as opposed to propagating them. Though constants of integration aren’t always added, as we’ll see in integrating factor.

We are not done yet. It passes through $(\pi, 0)$ , so

\begin{align*}\frac14 \sin(2\pi) + \frac12(\pi) &= \e^{\e^{0}} + E \\ 0 + \frac\pi2 &= \e^1 + E \\ \frac\pi2 - \e &= E \end{align*}

Final answer:

\frac14 \sin2x + \frac x2 = \e^{\e^y} + \frac\pi2 - \e \qed

The original markscheme accepted equivalent forms, though often it will ask you to evaluate it at a particular value of $x$ , just to make the grading easier.

Tips

\frac{\d y}{\d x} = f(x)g(y) \iff \int f(x)\d x = \int \frac{1}{g(y)}\d y