H7, H8 - Separable differential equations and models | AQA A-Level Mathematics

Solve simple first-order differential equations by separating variables and using initial conditions (H7). Then interpret cooling and motion models, checking solution domains and the assumptions that limit their predictions (H8).

Separating variables and retaining equilibria

A hot object cools quickly at first, then more slowly. If its cooling rate is proportional to its temperature above the surroundings, will it ever reach the surrounding temperature? A differential equation turns this rate statement into a model you can solve. Start with the simpler equation $dy/dx=x(y+2)$ : putting $y=-2$ makes both sides zero, while other solutions change with $x$ .

You already know

An antiderivative reverses differentiation. The supplied rule $\int f'(x)/f(x)\,dx=\ln|f(x)|+C$ handles logarithmic integrals; powers and exponentials use the integration rules you have learnt.

A first-order differential equation relates a function to its first derivative. It is separable if it can be written $dy/dx=f(x)g(y)$ . For example, $x(y+2)$ separates into an $x$ -factor and a $y$ -factor. A sum such as $x+y$ is not already in that form.

Separation of variables

Where $g(y)\ne0$ ,

\frac{dy}{dx}=f(x)g(y) \quad\Rightarrow\quad \int\frac1{g(y)}\,dy=\int f(x)\,dx

Use one arbitrary constant after integrating. Check zeros of $g(y)$ separately in the original equation before dividing.
Not in the formula booklet: learn it.

An equilibrium solution is a constant solution: its derivative is zero everywhere on its interval. A zero of the divided-out factor may supply one. A value making the original differential equation undefined supplies none. A general solution contains an arbitrary constant and describes a family of functions satisfying the equation.

Worked example 1: factor before dividing

Question. Find the general solution of $dy/dx=xy+2x$ , including constant solutions.

Plan. Factor out $x$ , check the zero of the other factor, then separate and integrate.

\begin{aligned} \frac{dy}{dx}&=x(y+2) && \text{common factor} \\ y=-2&\Rightarrow\frac{dy}{dx}=0=x(-2+2) && \text{check equilibrium} \\ \int\frac1{y+2}\,dy&=\int x\,dx && y\ne-2 \\ \ln|y+2|&=\frac12x^2+C && \text{integrate} \\ y+2&=Ae^{x^2/2} && \text{absorb the fixed sign} \end{aligned}

Why this step? Exponentiating first gives $|y+2|=e^Ce^{x^2/2}$ . The constant $A$ must allow either sign, because the logarithm described branches above and below $-2$ .

Answer. $y=Ae^{x^2/2}-2$ , for $A\in\mathbb R$ . Allowing $A=0$ restores the separately checked equilibrium. The family is defined for every real $x$ .

Check. $dy/dx=Ax e^{x^2/2}=x(y+2)$ , including $A=0$ .

Change only the constant in the $y$ -factor, from $2$ to $3$ . The equilibrium moves, while the separation process stays the same.

Worked example 2: move the equilibrium

Question. Find the general solution of $dy/dx=xy+3x$ , including constant solutions.

Plan. Factor, check $y=-3$ , and integrate the non-equilibrium branches.

\begin{aligned} \frac{dy}{dx}&=x(y+3) && \text{factorise} \\ y=-3&\Rightarrow0=x(-3+3) && \text{constant solution} \\ \ln|y+3|&=\frac12x^2+C && \text{separate and integrate} \\ y&=Ae^{x^2/2}-3 && \text{signed exponential constant} \end{aligned}

Why this step? The factor you divide by determines the excluded equilibrium; check the new factor rather than carrying over the old equilibrium.

Answer. $y=Ae^{x^2/2}-3$ , with $A\in\mathbb R$ , for all real $x$ .

Check. Its derivative is $Ax e^{x^2/2}=x(y+3)$ ; setting $A=0$ gives the original equation's zero derivative.

These solution families are derived by separation and are not printed in the formula booklet. Writing $A\in\mathbb R$ includes negative branches as well as the constant solution.

Watch out

Wrong: $y+2=e^Ce^{x^2/2}>0$ , so only solutions above $-2$ exist. Exponentiation initially gives an absolute value. Replace it with a signed nonzero constant for the changing solutions, then restore $A=0$ after checking the equilibrium in the original equation.

Move only the constant once more in your turn. Include any solution that the division would exclude.

Initial conditions, branches and solution domains

An initial condition, such as $y(0)=1$ , fixes the general solution's arbitrary constant and selects a particular solution. The resulting expression must satisfy both the original equation and the condition, on a connected interval containing the initial point.

For example, $1/(1-x^2)$ is undefined at $x=\pm1$ . A solution beginning at $x=0$ cannot pass through either pole to reach the other algebraic branches. The initial condition chooses an interval as well as a constant. No general- or particular-solution formula is supplied in the booklet; derive it from the differential equation.

Worked example 3: find a constant and an interval

Question. Solve $dy/dx=2xy^2$ , with $y(0)=1$ . State its largest connected solution interval containing $0$ .

Plan. Separate the nonzero branch, apply the initial condition and locate poles.

\begin{aligned} \int y^{-2}\,dy&=\int2x\,dx && y\ne0 \\ -\frac1y&=x^2+C && \text{power rule} \\ -1&=C && x=0,\ y=1 \\ y&=\frac1{1-x^2} && \text{rearrange} \end{aligned}

Why this step? The equilibrium $y=0$ solves the original equation but fails $y(0)=1$ , so division does not discard the required solution.

Answer. $y=1/(1-x^2)$ , for $-1<x<1$ .

Check. At $x=0$ , $y=1$ . Also $dy/dx=2x/(1-x^2)^2=2xy^2$ . Poles at $\pm1$ prevent extending this branch through either endpoint.

Change only the initial value from $1$ to $2$ . The differential equation is unchanged, but the constant changes the pole locations.

Worked example 4: a new initial value moves the poles

Question. Solve $dy/dx=2xy^2$ , with $y(0)=2$ . State its largest connected solution interval containing $0$ .

Plan. Use the same integrated equation with the new initial condition.

\begin{aligned} -\frac1y&=x^2+C && \text{separate and integrate} \\ -\frac12&=C && x=0,\ y=2 \\ y&=\frac2{1-2x^2} && \text{solve for }y \\ 1-2x^2&=0\Rightarrow x=\pm\frac1{\sqrt2} && \text{locate endpoints} \end{aligned}

Why this step? The solution must remain on the branch containing its initial point, rather than on the union of intervals where its formula happens to be defined.

Answer. $y=2/(1-2x^2)$ , for $-1/\sqrt2<x<1/\sqrt2$ .

Check. $y(0)=2$ and $dy/dx=8x/(1-2x^2)^2=2xy^2$ on this interval.

An initial value may also select a square-root sign. For $dy/dx=x/y$ , $y(0)=-2$ , integrate $y\,dy=x\,dx$ to obtain $y^2=x^2+4$ . Only $y=-\sqrt{x^2+4}$ meets the condition. It never vanishes, so the original equation is defined for all real $x$ ; its derivative is $-x/\sqrt{x^2+4}=x/y$ . The choice of sign cannot switch through $y=0$ , where the equation is undefined.

Watch out

Wrong: the domain in Worked example 3 is $\mathbb R\setminus\{-1,1\}$ as one initial-value solution. That set contains three separate branches. The largest connected interval containing the given initial point is $(-1,1)$ .

Change only the initial value again in the next problem. Apply it before deciding where the denominator vanishes.

Separation with partial fractions

After separating, you may need partial fractions to integrate the dependent-variable side. For $dy/dx=y(3-y)$ , first record the equilibria $y=0$ and $y=3$ . Dividing by $y(3-y)$ excludes both, so neither can be recovered from that divided equation alone.

The identity

\frac1{y(3-y)}=\frac13\left(\frac1y+\frac1{3-y}\right)

comes from recombining the right side. Its integral is $(\ln|y|-\ln|3-y|)/3$ , because the derivative of $3-y$ is $-1$ . This identity and the resulting solution are not in the formula booklet; derive them. The logarithmic derivative-ratio integral is supplied.

Worked example 5: a solution between two equilibria

Question. Solve $dy/dx=y(3-y)$ , with $y(0)=1$ .

Plan. Separate, integrate using partial fractions and use the branch between $0$ and $3$ .

\begin{aligned} \int\frac1{y(3-y)}\,dy&=\int1\,dx && y\ne0,3 \\ \ln\left|\frac y{3-y}\right|&=3x+C && \text{integrate and multiply by }3 \\ C&=\ln\frac12 && x=0,\ y=1 \\ \frac y{3-y}&=\frac12e^{3x} && 0<y<3 \\ y&=\frac3{1+2e^{-3x}} && \text{solve for }y \end{aligned}

Why this step? The initial value lies between the equilibria, so the ratio is positive. The final expression must stay on that branch.

Answer. $y=3/(1+2e^{-3x})$ , for every real $x$ .

Check. $y(0)=1$ , and $y'=18e^{-3x}/(1+2e^{-3x})^2=y(3-y)$ . The denominator is positive and $0<y<3$ throughout.

Change only the initial value from $1$ to $2$ . The partial-fraction integral stays the same; its integration constant changes.

Worked example 6: another branch within the same bounds

Question. Solve $dy/dx=y(3-y)$ , with $y(0)=2$ .

Plan. Reuse the integrated relationship and fit its constant to the new initial value.

\begin{aligned} \ln\left(\frac y{3-y}\right)&=3x+C && 0<y<3 \\ C&=\ln2 && x=0,\ y=2 \\ \frac y{3-y}&=2e^{3x} && \text{exponentiate} \\ y&=\frac3{1+\frac12e^{-3x}} && \text{rearrange} \end{aligned}

Why this step? The ratio at the initial point is now $2/(3-2)=2$ , rather than $1/2$ ; its reciprocal appears in the final denominator.

Answer. $y=3/(1+\frac12e^{-3x})$ , for every real $x$ .

Check. At $0$ this is $2$ . Its derivative is $(9/2)e^{-3x}/(1+\frac12e^{-3x})^2=y(3-y)$ , and it remains between the equilibria.

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H7, H8 - Separable differential equations and models

Separating variables and retaining equilibria

Initial conditions, branches and solution domains

Separation with partial fractions

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