4.10a - General and particular solutions | OCR A-Level Further Mathematics A

A differential equation describes a relationship between a function and its derivatives. You will distinguish a whole family of solutions from the particular member selected by initial or boundary conditions, and check both the equation and the conditions.

Families of solutions

If every point on a curve has gradient $2x$ , which curve is it? The curves $y=x^2$ , $y=x^2+3$ and $y=x^2-5$ all fit. Knowing every gradient still does not tell you the height of the curve.

You already know

Integrating $\dfrac{dy}{dx}=f(x)$ gives $y=\int f(x)\,dx+C$ . Differentiating a constant gives zero, so integration cannot recover its value without more information.

Here the equation fixes the shape but allows vertical translations. The diagram shows a family, with one member selected by a point.

[DIAGRAM: asset_name: A condition selects one solution; asset_slug: h245_4_10_solution_family; caption: Every curve has gradient 2x, but only y = x squared + 3 passes through the point (1, 4); description: Cartesian plot of y = x squared plus C for C equal to -3, 0, 3 and 6. The C = 3 curve is blue and the others grey. The point (1, 4) is marked and labelled, with a note that it selects C = 3.]

General and particular solutions

A general solution describes a family of solutions using arbitrary constants. A particular solution has those constants fixed by specified conditions.

For $y'=2x$ , the general solution is $y=x^2+C$ , where $C$ is an arbitrary real constant. The particular solution through $(1,4)$ is $y=x^2+3$ .

An arbitrary constant is fixed along any one solution, but may take different values for different members of the family. It need not appear as an added constant: $y=Ce^{2x}$ is a family too. Taking $C=0$ gives the valid solution $y=0$ of $y'=2y$ .

The formula booklet does not give these solution definitions or the procedures for verifying a solution and using conditions to determine its constants; learn these principles.

For the regular first and second order equations used here, a general solution has respectively one and two independent arbitrary constants. For example, integrating $y''=2$ twice gives $y=x^2+Ax+B$ . The order is the highest derivative present. Two symbols are not necessarily two independent constants: $y=x^2+A+B$ still has only one adjustable constant, $A+B$ .

There is a useful surprise here: the equation $y'=2x$ knows the gradient at every value of $x$ , yet cannot distinguish any of the infinitely many curves in the picture. One point supplies the missing information.

Key idea

The differential equation fixes a family; extra conditions select its member, if they are consistent and sufficient.

Keep the number of independent choices separate from the number of letters you have written.

Verifying a solution

A proposed solution must satisfy the differential equation throughout its stated interval, not just at the point used to find a constant. Differentiate first and substitute into the original equation.

Solution on an interval

A function is a solution on an interval when the required derivatives exist and substitution makes the differential equation true at every point of that interval. For example, $y=x^2+C$ solves $y'=2x$ on the whole real line; $y=x^2+x$ does not, since $y'=2x+1$ .

You can verify a supplied family without first knowing how someone found it. This is especially useful when its formula looks unfamiliar.

Worked example 1: checking a first order family

Question. Verify that $y=2+Ce^{-3x}$ solves $y'+3y=6$ for every real constant $C$ .

Plan. Differentiate the proposed solution, then calculate the whole left-hand side.

\begin{aligned} y'&=-3Ce^{-3x} &&\text{differentiate}\\ y'+3y&=-3Ce^{-3x}+3(2+Ce^{-3x}) &&\text{substitute}\\ &=6 &&\text{exponential terms cancel} \end{aligned}

Why this step? Substitution tests the equation itself; merely finding a derivative does not show that it satisfies the equation.

Answer. The family satisfies $y'+3y=6$ for all real $x$ and every real $C$ .

Check. When $C=0$ , the constant solution $y=2$ gives $0+3(2)=6$ .

Change just the order of the equation in the next example. You now need a second derivative, but the test is still substitution into the complete left-hand side.

Worked example 2: checking a second order family

Question. Verify that $y=Ae^x+Be^{-2x}$ solves $y''+y'-2y=0$ .

Plan. Find both derivatives, then collect the coefficients of each exponential separately.

\begin{aligned} y'&=Ae^x-2Be^{-2x} &&\text{first derivative}\\ y''&=Ae^x+4Be^{-2x} &&\text{second derivative}\\ y''+y'-2y&=(1+1-2)Ae^x &&\text{collect }e^x\\ &\quad +(4-2-2)Be^{-2x} &&\text{collect }e^{-2x}\\ &=0 &&\text{both coefficients vanish} \end{aligned}

Why this step? The independent exponential terms must cancel separately, whatever $A$ and $B$ are.

Answer. The family satisfies the equation for every real $x$ .

Check. Choosing $A=0$ , $B=1$ gives $4e^{-2x}-2e^{-2x}-2e^{-2x}=0$ .

Passing a condition alone is much weaker than passing the equation: many curves pass through one point.

Watch out

Wrong: “ $y(0)=2$ , so this solves the differential equation.” Right: check the derivative equation for general $x$ , and check the condition separately.

Use the same cancellation test with different exponents.

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4.10a - General and particular solutions

Families of solutions

Verifying a solution

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