1.08k, 1.08l - Separable differential equations and their solutions | OCR A-Level Mathematics A

Find general and particular solutions of first-order differential equations by direct integration or separation of variables. You will keep constant solutions, select valid branches and interpret rates, limits and restrictions in a motion model.

Solutions and initial information

Can a falling object keep speeding up without ever reaching $12\,\mathrm{m\,s^{-1}}$ ? A supplied model gives its downward velocity $v$ by $\mathrm dv/\mathrm dt=6-v/2$ , starting from rest. At $v=0$ , its acceleration is $6$ ; at $v=6$ , it is $3$ ; at $v=9$ , it is only $1.5$ , all in $\mathrm{m\,s^{-2}}$ . Solving the equation will tell us whether the speed reaches $12$ and when it reaches halfway.

You already know

Integration reverses differentiation. An indefinite integral needs an arbitrary constant. The chain rule differentiates a composite function, and $\int 1/x\,\mathrm dx=\ln|x|+C$ for $x\ne0$ .

A solution must satisfy the rate rule throughout its interval, not merely pass through one point.

General and particular solutions

A solution is a differentiable function satisfying the differential equation. A general solution contains an arbitrary constant describing a family of solutions. An initial condition, such as $y(-1)=1$ , can select a particular solution. For example, every $y=x^2+C$ solves $\mathrm dy/\mathrm dx=2x$ , but only $C=3$ gives $y(0)=3$ .

If the derivative depends only on the independent variable, integrate directly. This solution method is not in the formula sheet: learn it.

Worked example 1: a point selects a curve

Question. Solve $\mathrm dy/\mathrm dx=3x^2+2$ , given $y(-1)=1$ .

Plan. Integrate with a constant, then substitute both coordinates of the initial point.

\begin{aligned} y&=x^3+2x+C && \text{integrate each term} \\ 1&=(-1)^3+2(-1)+C && \text{use }y(-1)=1 \\ C&=4 && \text{solve for the constant} \end{aligned}

Why this step? The given ordinate is $1$ , but the other terms at $x=-1$ contribute $-3$ , so $C$ is not $1$ .

Answer. $y=x^3+2x+4$ , for real $x$ .

Check. Differentiating gives $3x^2+2$ , and $y(-1)=-1-2+4=1$ .

Change only the input of the initial point from $-1$ to $1$ . The derivative stays the same, so the curves differ only by a vertical translation.

Worked example 2: a different initial input

Question. Solve $\mathrm dy/\mathrm dx=3x^2+2$ , given $y(1)=1$ .

Plan. Use the same general solution and fit its constant to the new point.

\begin{aligned} y&=x^3+2x+C && \text{general solution} \\ 1&=1+2+C && \text{substitute }x=1 \\ C&=-2 && \text{determine the translation} \end{aligned}

Why this step? A point condition selects the member of the family; it does not change the differential equation.

Answer. $y=x^3+2x-2$ .

Check. The derivative is still $3x^2+2$ , and $y(1)=1+2-2=1$ .

A domain can split a family into separate intervals. For $\mathrm dy/\mathrm dx=1/(x-2)$ , direct integration gives $y=\ln|x-2|+C$ . The equation is undefined at $2$ , so a condition at $x=3$ selects an interval on the right, never a solution passing through the pole.

Watch out

Wrong: take the initial ordinate as $C$ without substitution. Right: substitute the entire point into the integrated equation. Then check both the derivative and the point condition.

Move the initial input once more in the next problem.

Separating variables before integration

When the right-hand side involves both variables, first look for a product of a function of $x$ and a function of $y$ . Factorisation may be the step that reveals it.

Separation of variables

\frac{\mathrm dy}{\mathrm dx}=g(x)h(y),

then, where $h(y)\ne0$ , separate and integrate:

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

Check constant solutions from $h(y)=0$ separately, and restrict the solution to an interval where the equation is defined.

Not in the formula sheet: learn it.

The notation is backed by the chain rule: differentiating the left integral with respect to $x$ produces $(1/h(y))\mathrm dy/\mathrm dx$ . It is not permission to treat a varying $y$ as constant in an integral with respect to $x$ .

Worked example 3: factor before separating

Question. Solve $\mathrm dy/\mathrm dx=xy+y$ , given $y(0)=2$ .

Plan. Factor out $y$ , separate the variables, then use the initial condition after integrating.

\begin{aligned} \frac{\mathrm dy}{\mathrm dx}&=(x+1)y && \text{common factor} \\ \int\frac1y\,\mathrm dy&=\int(x+1)\,\mathrm dx && y\ne0 \\ \ln|y|&=\frac{x^2}{2}+x+C && \text{integrate} \\ y&=Ae^{x^2/2+x} && \text{absorb the sign into }A \\ 2&=Ae^0\quad\Longrightarrow\quad A=2 && \text{initial condition} \end{aligned}

Why this step? Exponentiating converts an additive logarithmic constant into a multiplicative constant. The positive initial value selects a positive branch.

Answer. $y=2e^{x^2/2+x}$ , for real $x$ . The constant solution $y=0$ , excluded during division, does not meet this initial condition.

Check. Differentiation gives $y'=(x+1)y$ , and $y(0)=2$ .

Change only the dependent-variable factor from $y$ to $y-1$ . Separate the whole factor; the quantity inside the logarithm changes with it.

Worked example 4: a shifted dependent variable

Question. Solve $\mathrm dy/\mathrm dx=xy+y-x-1$ , given $y(0)=2$ .

Plan. Factor into $(x+1)(y-1)$ , then integrate the reciprocal of $y-1$ .

\begin{aligned} \frac{\mathrm dy}{\mathrm dx}&=(x+1)(y-1) && \text{group and factor} \\ \int\frac1{y-1}\,\mathrm dy&=\int(x+1)\,\mathrm dx && y\ne1 \\ \ln|y-1|&=\frac{x^2}{2}+x+C && \text{integrate} \\ y-1&=Ae^{x^2/2+x} && \text{exponentiate} \\ 2-1&=A\quad\Longrightarrow\quad A=1 && \text{fit the point} \end{aligned}

Why this step? The common factor is $y-1$ , so dividing by $y$ would not separate the equation.

Answer. $y=1+e^{x^2/2+x}$ , for real $x$ . The separate equilibrium $y=1$ fails the initial condition.

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

The same calculation solves proportional decay. For $P'=-0.2P$ , $P(0)=50$ , separation gives $\ln P=-0.2t+C$ , then $P=50e^{-0.2t}$ . Its derivative is negative, but its magnitude decreases as the positive quantity gets smaller. These exponential solutions are derived here; they are not printed in the formula sheet.

Watch out

Wrong: $y=e^{x^2/2+x}+C$ after $\ln|y|=x^2/2+x+C$ . Right: $y=Ae^{x^2/2+x}$ . A real signed $A$ includes the branch sign; check $A=0$ in the equation before including it.

For the next shifted equation, change the scale of the right-hand side and the initial value. Show the separation rather than only writing a remembered exponential.

Recovering equilibria and choosing a valid branch

Division can erase a perfectly good constant solution. Record the zero factors first, and use the initial value to choose the relevant branch afterwards.

Equilibrium solution

An equilibrium solution is a constant function satisfying the differential equation. For $y'=y(2-y)$ , both $y=0$ and $y=2$ have derivative zero and make the right-hand side zero. A constant such as $y=1$ is not an equilibrium: its derivative is zero but the right-hand side is $1$ .

Partial fractions from your earlier integration work handle a product of two linear factors. Keep the minus sign from integrating $1/(2-y)$ .

Worked example 5: keep both constant solutions

Question. Identify the equilibria of $y'=y(2-y)$ , then solve with $y(0)=1$ .

Plan. Check the zero factors, separate the nonconstant solutions and use partial fractions.

\begin{aligned} y(2-y)&=0\quad\Longrightarrow\quad y=0\text{ or }y=2 && \text{equilibria} \\ \frac1{y(2-y)}&=\frac1{2y}+\frac1{2(2-y)} && \text{partial fractions} \\ \frac12\ln|y|-\frac12\ln|2-y|&=x+C && \text{separate and integrate} \\ \frac{y}{2-y}&=Ke^{2x} && \text{combine logarithms} \\ 1&=K\quad\Longrightarrow\quad y=\frac2{1+e^{-2x}} && \text{use }y(0)=1 \end{aligned}

Why this step? Division excluded both equilibria. The logarithmic relation describes only the other solutions, so it cannot by itself be the complete answer about equilibria.

Answer. The equilibria are $0$ and $2$ . The particular solution is $y=2/(1+e^{-2x})$ , for real $x$ .

Check. With $z=e^{-2x}>0$ , $y'=4z/(1+z)^2=y(2-y)$ , and $y(0)=1$ . Also $0<y<2$ .

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Pure Mathematics

Statistics

Mechanics

1.08k, 1.08l - Separable differential equations and their solutions

Solutions and initial information

Separating variables before integration

Recovering equilibria and choosing a valid branch

Continue this lesson.

Learn faster with Nova