1.1 Separable ODEs

A separable first-order ODE can be written in the form

equation (1.1) (1.1)

y^{\prime}=g(x)h(y).

We will assume that $g(x)$ is continuous over some range of values of $x$ , e.g. $x\in(a,b)$ , possibly the whole real line $(-\infty,\infty)$ . Continuity of $g(x)$ ensures that we can take the necessary integrals.

1.1.1 A simple solution

Any (constant) solution of the equation $h(y)=0$ is also a possible solution of (1.1).

Example 1.1.

Show that $y=n\pi$ ( $n\in\mathbb{Z}$ ) is a possible solution of the first-order ODE,

y^{\prime}=x\sin(y).

Solution.

Here $g(x)=x$ and $h(y)=\sin(y)$ . Now, $h(y)=\sin(y)=0$ if $y=n\pi$ (where $n\in\mathbb{Z}$ ), so the right-hand side of the ODE is zero. Clearly $y=n\pi$ (constant) $\forall x$ $\Longrightarrow$ $y^{\prime}=0$ (i.e. the left-hand side is also zero). Therefore $y=n\pi$ is a possible solution.

1.1.2 Separation of variables

To derive a more general solution, we must assume that $h(y)\neq 0$ , so that we can divide each side by $h(y)$ ,

\frac{y^{\prime}}{h(y)}=g(x).

Integrating each side with respect to $x$ ,

\int\frac{y^{\prime}}{h(y)}~{}{\rm d}x=\int g(x)~{}{\rm d}x.

since $\mathrm{d}y=y^{\prime}\,\mathrm{d}x$ , this implies that

equation (1.2) (1.2)

\int\frac{{\rm d}y}{h(y)}=\int g(x)~{}{\rm d}x.

Evaluating these integrals leads to a solution for $y(x)$ . Sometimes the general solution will contain solutions of $h(y)=0$ , but this is not always the case.

Example 1.2.

Find the general solution to

y^{\prime}=x(y+1)^{2}.

Solution.

Clearly $y=-1$ is a possible constant solution. If $y\neq-1$ , we separate variables,

\displaystyle\int\frac{{\rm d}y}{(y+1)^{2}}

\displaystyle=

\displaystyle\int x~{}{\rm d}x

\displaystyle\Longrightarrow-\frac{1}{y+1}

\displaystyle=

\displaystyle\frac{x^{2}}{2}+C,

where C is an arbitrary constant of integration. Rearranging this expression,

\displaystyle-\frac{2}{y+1}

\displaystyle=

\displaystyle x^{2}+2C

\displaystyle\Longrightarrow y+1

\displaystyle=

\displaystyle-\frac{2}{x^{2}+2C}

\displaystyle\Longrightarrow y

\displaystyle=

\displaystyle-1-\left(\frac{2}{x^{2}+D}\right),

where $D=2C$ is a rescaled constant. Alongside the special solution $y=-1$ , this is the most general solution to this first order ODE. Different values of $D$ lead to a family of solution curves (or integral curves) in the $xy$ -plane.

Check (not lectured): Differentiating this function, we see that

y^{\prime}=\frac{4x}{\left(x^{2}+D\right)^{2}}=x(y+1)^{2},

as required.

Example 1.3.

Find the general solution of the ODE $y^{\prime}=(y+xy)/x^{2}$ and then the particular solution which passes through the point $(1,1)$ , i.e. $y(1)=1$ .

Solution.

We have

\displaystyle y^{\prime}

\displaystyle=

\displaystyle\frac{(y+xy)}{x^{2}}

\displaystyle=

\displaystyle\frac{(1+x)y}{x^{2}}.

Clearly $y=0$ is a possible solution, but this is incompatible with $y(1)=1$ so can be ignored. Assuming $y\neq 0$ , separation of variables leads to

\int\frac{{\rm d}y}{y}=\int\left(\frac{1+x}{x^{2}}\right){\rm d}x=\int\left(% \frac{1}{x^{2}}+\frac{1}{x}\right)\,\mathrm{d}x\Longrightarrow\ln|y|=-\frac{1}% {x}+\ln|x|+C

(where $C$ is the constant of integration). Taking the exponential of both sides,

|y|=\exp\left[-\frac{1}{x}+\ln|x|+C\right]=|x|\exp\left[-\frac{1}{x}+C\right]=% |x|\exp[C]\exp\left[-\frac{1}{x}\right],

where $\displaystyle{\exp(x)\equiv\mathrm{e}^{x}}$ . Dropping the modulus operators from both sides,

y=\pm\,x\exp[C]\exp\left[-\frac{1}{x}\right].

Absorbing the plus/minus sign into a new constant $D$ (where $|D|=\exp[C]$ ),

y=Dx\exp\left[-\frac{1}{x}\right].

If $y(1)=1$ , this implies that

1=D\cdot 1\cdot\exp[-1]\equiv D\mathrm{e}^{-1}\Rightarrow D=\mathrm{e}.

So the required solution is,

y=\mathrm{e}x\exp\left[-\frac{1}{x}\right]=x\exp\left[1-\frac{1}{x}\right]% \left(\equiv x\mathrm{e}^{\left[1-\left(1/x\right)\right]}\right).

Example 1.4.

If $N(t)$ is the size of a radioactive sample at time $t$ , then

\frac{\mathrm{d}N}{\mathrm{d}t}=-\alpha N,

where $\alpha$ is a positive constant. How long does it take for half the sample to decay?

Solution.

Assuming that $N\neq 0$ , we again proceed by separating variables (note the change of notation!):

\int\frac{{\rm d}N}{N}=-\alpha\int\,\mathrm{d}t\Longrightarrow\ln|N|=-\alpha t% +C,

where $C$ is a constant of integration. Hence,

|N(t)|=\mathrm{e}^{C}\mathrm{e}^{-\alpha t}\Longrightarrow N(t)=\pm\,\mathrm{e% }^{C}\mathrm{e}^{-\alpha t}\Longrightarrow N(t)=A\mathrm{e}^{-\alpha t},

where $A$ is a new constant ( $|A|=\mathrm{e}^{C}$ ) into which the plus/minus sign has been absorbed. If $N(0)=N_{0}$ (initial sample size) then,

N_{0}=A\mathrm{e}^{0}=A\Longrightarrow N(t)=N_{0}\mathrm{e}^{-\alpha t}.

Let $t_{h}$ be the length of time it takes for half the sample to decay, i.e. $N(t_{h})=N_{0}/2$ . This implies that,

\frac{N_{0}}{2}=N(t_{h})=N_{0}\mathrm{e}^{-\alpha t_{h}}.

Hence

\displaystyle\frac{1}{2}

\displaystyle=

\displaystyle\mathrm{e}^{-\alpha t_{h}}

\displaystyle\mbox{(taking logs)}\;\Longrightarrow\ln\left[\frac{1}{2}\right]

\displaystyle=

\displaystyle-\alpha t_{h}

\displaystyle\Longrightarrow\ln 1-\ln 2

\displaystyle=

\displaystyle-\alpha t_{h}\quad\mbox{(recall that $\ln(x/y)=\ln x-\ln y)$}

\displaystyle\Longrightarrow 0-\ln 2

\displaystyle=

\displaystyle-\alpha t_{h}

\displaystyle\Longrightarrow t_{h}

\displaystyle=

\displaystyle\frac{\ln 2}{\alpha}.

This is the time it takes for half the sample to decay (known as the sample’s half-life).