3.8 - Solving separable differential equations | Cambridge A-Level Mathematics

Knowing a rate of change can tell us how a quantity develops, even when we do not yet know its formula. In this part of 3.8, you will translate rate statements into differential equations, separate variables and integrate, then use an initial value to select a particular solution. Checking the original equation will also catch solutions lost during division.

Rates form equations

If a quantity grows faster when there is more of it, what does that tell us about its formula? Before finding the formula, we can express the relationship between the quantity and its rate.

If $M$ depends on time $t$ , its instantaneous rate of change is $\mathrm dM/\mathrm dt$ . A differential equation relates an unknown function to one or more of its derivatives. Here we use first-order equations, involving only a first derivative.

Proportional rate

For a positive quantity $M$ , “the rate of increase of $M$ is proportional to $M$ ” means

\frac{\mathrm dM}{\mathrm dt}=kM

for a constant of proportionality $k>0$ . A rate of decrease of this size gives $\mathrm dM/\mathrm dt=-kM$ . The derivative is signed; the stated size of a decrease is positive.

A measurement of both the quantity and its rate determines $k$ . Keep it constant after evaluating it. More generally, “proportional to the product of $A$ and $B$ ” means $kAB$ , while “inversely proportional to $A$ ” means $k/A$ , on a domain where that expression is defined.

Worked example 1: increasing mass

Question. A mass $M$ , measured in grams, increases at a rate proportional to its present value. Time $t$ is measured in minutes. When $M=40$ , its rate of increase is $6\,\mathrm{g\,min^{-1}}$ . Form a differential equation for $M$ .

Plan. Write the proportional law with an unknown constant, then substitute the measured mass and rate.

\frac{\mathrm dM}{\mathrm dt}=kM, \qquad6=40k\quad\Longrightarrow\quad k=0.15.

Answer. $\mathrm dM/\mathrm dt=0.15M$ , with $k=0.15\,\mathrm{min^{-1}}$ .

Check. At $M=40$ , the equation gives $0.15\times40=6$ . Units are $\mathrm{min^{-1}}\times\mathrm g=\mathrm{g\,min^{-1}}$ .

Change only increase to decrease. The magnitude of the proportionality constant is unchanged, but the derivative changes sign.

Worked example 2: decreasing mass

Question. A mass $M$ , in grams, decreases at a rate proportional to its present value. When $M=40$ , the rate of decrease is $6\,\mathrm{g\,min^{-1}}$ . Form a differential equation, with $t$ in minutes.

Plan. Use a negative derivative for a decreasing quantity, and take $k>0$ for the size of the proportionality constant.

\frac{\mathrm dM}{\mathrm dt}=-kM, \qquad-6=-40k\quad\Longrightarrow\quad k=0.15.

Answer. $\mathrm dM/\mathrm dt=-0.15M$ .

Check. At $M=40$ , the derivative is $-6$ , a loss of $6$ grams per minute. At a positive mass the derivative remains negative, as the description requires.

The constant can be written with a sign instead: $\mathrm dM/\mathrm dt=cM$ with $c=-0.15$ . Both conventions give the same equation; do not include the negative sign twice.

Watch out

“Proportional to” does not mean “equal to”. Write the constant first, then evaluate it from the data. For a decrease of $6$ per minute, the derivative is $-6$ , not $6$ .

Form a new signed rate equation.

Separating and integrating

An equation such as $\mathrm dy/\mathrm dx=2xy$ contains an unknown function and its derivative. When the right side is a product of a function of $x$ and a function of $y$ , we can collect the $y$ -terms on one side and the $x$ -terms on the other.

Separable equation

A first-order equation is separable when it can be written

\frac{\mathrm dy}{\mathrm dx}=f(x)g(y).

Where $g(y)\ne0$ , separate it as

\frac1{g(y)}\,\mathrm dy=f(x)\,\mathrm dx

and integrate both sides. Before dividing, test any constant values for which $g(y)=0$ in the original equation.

This notation follows the chain rule: if $G'(y)=1/g(y)$ , then $\mathrm d[G(y(x))]/\mathrm dx=(1/g(y))\mathrm dy/\mathrm dx=f(x)$ . Integration therefore gives $G(y)=\int f(x)\,\mathrm dx+C$ . One arbitrary constant is enough because the difference of two integration constants is another arbitrary constant.

Worked example 3: a logarithm on separation

Question. Find the general solution of $\mathrm dy/\mathrm dx=2xy$ .

Plan. Check the constant solution $y=0$ , then divide by $y$ and integrate.

For $y\ne0$ ,

\begin{aligned} \frac1y\,\mathrm dy&=2x\,\mathrm dx\\ \ln|y|&=x^2+C\\ y&=Ae^{x^2} \end{aligned}

For the nonzero solutions, $A$ is any nonzero real constant: the sign of $y$ is absorbed in it. The constant solution $y=0$ is recovered by also allowing $A=0$ .

Answer. $y=Ae^{x^2}$ , for any real constant $A$ .

Check. Differentiating gives $\mathrm dy/\mathrm dx=2xAe^{x^2}=2xy$ , including when $A=0$ .

Keep $2x$ , but replace the $y$ -factor by $y(1-y)$ . This adds a second constant solution and brings partial fractions into the separated integral.

Worked example 4: two constant solutions

Question. Solve $\mathrm dy/\mathrm dx=2xy(1-y)$ , including constant solutions.

Plan. Check $y=0$ and $y=1$ before dividing, then split the separated fraction.

Both constant values satisfy the original equation. Away from them,

\begin{aligned} \frac{1}{y(1-y)}\,\mathrm dy&=2x\,\mathrm dx\\ \frac{1}{y(1-y)}&=\frac1y+\frac1{1-y}\\ \ln|y|-\ln|1-y|&=x^2+C \end{aligned}

The negative sign before the second logarithm comes from integrating $1/(1-y)$ .

Combining the logarithms and exponentiating gives $y/(1-y)=Ke^{x^2}$ , with $K\ne0$ of either sign. Rearranging,

y=\frac{Ke^{x^2}}{1+Ke^{x^2}}.

Answer. This family, on intervals where its denominator is nonzero, together with $y=0$ and $y=1$ . Allowing $K=0$ includes $y=0$ , but no finite $K$ gives $y=1$ .

Check. Let $u=Ke^{x^2}$ . Then $y=u/(1+u)$ and $y'=2xu/(1+u)^2=2xy(1-y)$ . The two constant solutions also have zero derivative and zero right side.

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3.8 - Solving separable differential equations

Rates form equations

Separating and integrating

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