Mc32, Mc33 - Separable differential equations and their solutions | OCR A-Level Mathematics B (MEI)

Solve first-order differential equations analytically, using direct integration or separation of variables. You will use initial information, preserve equilibrium solutions, identify valid intervals and interpret a supplied motion model without extending it beyond its assumptions.

Solutions and initial information

Can one acceleration law describe an object speeding up and another slowing down towards the same velocity? Take the supplied model $\mathrm dv/\mathrm dt=5-v/4$ , with downward velocity positive. At $v=0$ , acceleration is $5$ ; at $v=30$ , it is $-2.5$ , in $\mathrm{m\,s^{-2}}$ . Both point towards $v=20$ . We will find the two motions and decide whether either reaches that value in finite time.

You already know

Direct integration recovers a function from its derivative. Integration by parts is $\int u\,\mathrm dv=uv-\int v\,\mathrm du$ , and is in the formula sheet. The chain rule checks a recovered composite function.

The equation determines the family; the initial information selects a member of that family.

General and particular solutions

A solution is a differentiable function satisfying the differential equation on its stated interval. A general solution includes an arbitrary constant. An initial condition can determine that constant and produce a particular solution. For example, $y=x^2+C$ solves $y'=2x$ , while $y(0)=3$ selects $y=x^2+3$ .

When the derivative is a function of the independent variable alone, direct integration is enough. This method is not in the formula sheet: learn it. The integral itself may require a method you already know.

Worked example 1: direct integration can involve parts

Question. Solve $\mathrm dy/\mathrm dx=(6x-3)e^{-x}$ , given $y(0)=5$ .

Plan. Integrate $6xe^{-x}$ by parts, integrate the remaining exponential, then use the initial condition.

For the first term, take $u=6x$ and $\mathrm dv=e^{-x}\,\mathrm dx$ , so $v=-e^{-x}$ .

\begin{aligned} \int6xe^{-x}\,\mathrm dx&=-6xe^{-x}+6\int e^{-x}\,\mathrm dx && \text{integration by parts} \\ &=-6xe^{-x}-6e^{-x} && \text{integrate the exponential} \\ y&=-6xe^{-x}-6e^{-x}+3e^{-x}+C && \text{include the }-3e^{-x}\text{ term} \\ &=-(6x+3)e^{-x}+C && \text{collect like terms} \\ 5&=-3+C\quad\Longrightarrow\quad C=8 && \text{initial condition} \end{aligned}

Why this step? Differentiating the linear factor simplifies it, so integration by parts removes the product without introducing a harder integral.

Answer. $y=8-(6x+3)e^{-x}$ , for real $x$ .

Check. $y'=-6e^{-x}+(6x+3)e^{-x}=(6x-3)e^{-x}$ , and $y(0)=8-3=5$ .

Change only the constant inside the derivative's bracket from $-3$ to $+3$ . The integral's constant coefficient changes sign at the earlier integration step, not at the end.

Worked example 2: one changed term

Question. Solve $\mathrm dy/\mathrm dx=(6x+3)e^{-x}$ , given $y(0)=5$ .

Plan. Reuse the parts calculation for $6xe^{-x}$ , then integrate the changed exponential term.

\begin{aligned} y&=-6xe^{-x}-6e^{-x}-3e^{-x}+C && \text{integrate both terms} \\ &=-(6x+9)e^{-x}+C && \text{collect coefficients} \\ 5&=-9+C\quad\Longrightarrow\quad C=14 && \text{fit the initial value} \end{aligned}

Why this step? The integral of $+3e^{-x}$ is $-3e^{-x}$ , so it adds to the negative exponential contribution from integration by parts.

Answer. $y=14-(6x+9)e^{-x}$ .

Check. $y'=-6e^{-x}+(6x+9)e^{-x}=(6x+3)e^{-x}$ , and $y(0)=14-9=5$ .

For a simpler direct example, $y'=3x^2+2$ gives $y=x^3+2x+C$ ; $y(-1)=1$ then gives $C=4$ . Fitting a point does not excuse an incorrect derivative: both must be checked.

Intervals also matter. The equation $y'=1/(x-2)$ gives $y=\ln|x-2|+C$ separately on $x<2$ and $x>2$ . An initial condition at $3$ selects the right-hand interval, without allowing the solution through the undefined input $2$ .

Watch out

Wrong: set $C$ equal to the initial ordinate before substituting the input. Right: put both coordinates into the entire integrated expression. In the second example, $C=14$ , although $y(0)=5$ .

Try a different linear factor and initial value. Keep the complete integration visible.

Separating variables before integration

An equation involving both $x$ and $y$ may still be simple after factorisation. Look for one factor containing only $x$ , multiplied by one containing only $y$ .

Separable equation

An equation is separable when it can be written $\mathrm dy/\mathrm dx=g(x)h(y)$ . For example, $xy+y=y(x+1)$ is a separable right-hand side. An unfactorised sum such as $x+y$ is not a product of that form.

Move the entire dependent-variable factor to the left before either integral is taken.

Separation of variables

Where $h(y)\ne0$ ,

\frac{\mathrm dy}{\mathrm dx}=g(x)h(y) \quad\Longrightarrow\quad \int\frac1{h(y)}\,\mathrm dy=\int g(x)\,\mathrm dx.

Check solutions from $h(y)=0$ separately. Integrate on an interval where the equation and the chosen solution are defined.

Not in the formula sheet: learn it.

This rearrangement reflects the chain rule: the derivative with respect to $x$ of the left integral is $y'/h(y)$ . It does not allow us to integrate with respect to $x$ while pretending a changing $y$ is constant.

Worked example 3: a common factor reveals the method

Question. Solve $y'=2xy+2x+2y+2$ , given $y(0)=2$ .

Plan. Factor the right side, separate $y+1$ and use the point condition after integration.

\begin{aligned} y'&=2(x+1)(y+1) && \text{factor by grouping} \\ \int\frac1{y+1}\,\mathrm dy&=\int2(x+1)\,\mathrm dx && y\ne-1 \\ \ln|y+1|&=x^2+2x+C && \text{integrate} \\ y+1&=Ae^{x^2+2x} && \text{use a signed constant} \\ 3&=A && \text{initial condition} \end{aligned}

Why this step? Exponentiating gives a multiplicative constant. Its sign must also record whether $y+1$ is positive or negative.

Answer. $y=3e^{x^2+2x}-1$ , for real $x$ . The equilibrium $y=-1$ does not satisfy the initial condition.

Check. $y'=6(x+1)e^{x^2+2x}=2(x+1)(y+1)$ , and $y(0)=2$ .

Change only the initial value to $-2$ . The solution now lies on the other side of the equilibrium, so the sign hidden by the absolute value becomes essential.

Worked example 4: the negative branch

Question. Solve $y'=2xy+2x+2y+2$ , given $y(0)=-2$ .

Plan. Use the separated family, retaining a real signed multiplier rather than assuming it is positive.

\begin{aligned} \ln|y+1|&=x^2+2x+C && \text{separation and integration} \\ y+1&=Ae^{x^2+2x} && \text{restore the branch sign} \\ -2+1&=A\quad\Longrightarrow\quad A=-1 && \text{use the new point} \\ y&=-e^{x^2+2x}-1 && \text{particular solution} \end{aligned}

Why this step? Writing $A=e^C>0$ would discard every solution below $y=-1$ . A signed constant includes the negative branch too.

Answer. $y=-e^{x^2+2x}-1$ , for real $x$ .

Check. $y(0)=-2$ ; differentiating gives $-2(x+1)e^{x^2+2x}=2(x+1)(y+1)$ .

For the unshifted equation $y'=xy+y$ , the same method gives $y=Ae^{x^2/2+x}$ . The condition $y(0)=2$ gives $A=2$ . Check $y=0$ separately: it satisfies the equation, even though the logarithmic step excludes it. Similarly, $P'=-0.2P$ , $P(0)=50$ , gives $P=50e^{-0.2t}$ , by integrating $1/P$ and fitting the multiplier. These derived solution families are not printed in the formula sheet.

Exam tip

Show the separated equation before integrating when detailed reasoning is requested. After removing logarithms, use a multiplicative constant: an added $+C$ outside the exponential is generally wrong.

The next equation changes the common coefficient and the initial value. Start by grouping its four terms.

Recovering equilibria and choosing a valid branch

The interval of a solution can be narrower than the domain of its differential equation. A denominator that appears while solving may create a finite-time pole.

Equilibrium solution

An equilibrium solution is a constant function satisfying the equation. For $y'=y(2-y)$ , $y=0$ and $y=2$ are equilibria because both sides are zero. Dividing by $y(2-y)$ excludes them from the subsequent calculation, rather than proving they do not exist.

Use partial fractions to integrate the other solutions, then fit the branch and interval to the initial information.

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Mc32, Mc33 - Separable differential equations and their solutions

Solutions and initial information

Separating variables before integration

Recovering equilibria and choosing a valid branch

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