Solving Second Order Differential Equations (Edexcel A Level Further Maths: Core Pure)

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Roger

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15 March 2024

Second Order Differential Equations

What is a second order differential equation?

A second order differential equation is an equation containing second order derivatives (and possibly first order derivatives) but no higher order derivatives

For example $\frac{d^{2} y}{d x^{2}} = 12 x$ is a second order differential equation
And so is $\frac{d^{2} x}{d t^{2}} - 5 \frac{d x}{d t} + 7 x = 5 \sin t$
But $\frac{d^{3} y}{d x^{3}} + 3 \frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} = 0$ is not, because it contains the third order derivative $\frac{d^{3} y}{d x^{3}}$

What are the types of second order differential equation?

We divide second order differential equations into two main types
A homogeneous second order differential equation is of the form $a \frac{d^{2} y}{d x^{2}} + b \frac{d y}{d x} + c y = 0$ where a, b and c are real constants

You may also see this written in the form $a y'' + b y' + c y = 0$ where $y'' = \frac{d^{2} y}{d x^{2}}$ and $y' = \frac{d y}{d x}$

A non-homogeneous second order differential equation is of the form $a \frac{d^{2} y}{d x^{2}} + b \frac{d y}{d x} + c y = f (x)$ where a, b and c are real constants and where f(x) is a non-zero function of x

You may also see this written in the form $a y'' + b y' + c y = f (x)$

How can I solve simple second order differential equations?

If a second order differential equation is of the form $\frac{d^{2} y}{d x^{2}} = f (x)$ , it will often be possible to solve it simply by using repeated integration

A separate integration constant will need to be included for each of the integrations

This means you will end up with two integration constants in your final answer
To find the values of these constants you will need two separate initial or boundary conditions

See the worked example below for examples of this

Worked example

Find the general solution to the second order differential equation

\frac{d^{2} y}{d x^{2}} = 0

8-2-1-2nd-order-diffes-a-we-solution

Find the particular solution to the second order differential equation

\frac{d^{2} y}{d x^{2}} = 24 x

that satisfies

\frac{d y}{d x} = - 2

and

y = 1

when

x = 0

8-2-1-2nd-order-diffes-b-we-solution

Auxiliary Equations & Complementary Functions

What is a complementary function?

For a second order homogeneous differential equation, the equation’s complementary function is the general solution to the equation

If the differential equation contains initial or boundary conditions you may then use those to find the precise solution to the equation

For a second order non-homogeneous differential equation, the equation’s complementary function is only a part of the general solution to the equation

For the complete general solution you will need to include the particular integral as well (see the following section)

In order to find a differential equation’s complementary function we use the associated auxiliary equation

What is an auxiliary equation?

For a second order differential equation $a \frac{d^{2} y}{d x^{2}} + b \frac{d y}{d x} + c y = f (x)$ the associated auxiliary equation is $a m^{2} + b m + c = 0$

It is possible that f(x)=0, in which case the differential equation is homogeneous
The auxiliary equation is exactly the same whether the differential equation is homogeneous or non-homogeneous
The auxiliary equation is a quadratic equation in the variable m

The solutions to the auxiliary equation will determine the nature of the associated complementary function

How do I use the auxiliary equation to find the associated complementary function?

STEP 1: Solve the auxiliary equation $a m^{2} + b m + c = 0$ to find its roots α and β

It is possible that the roots will be repeated, with α = β

STEP 2: The complementary function will be determined by the nature of the roots of the auxiliary equation:

CASE 1: $b^{2} - 4 a c > 0$ so that α and β are distinct real roots

The complementary function is $y = A e^{α x} + B e^{β x}$ where A and B are arbitrary constants
This holds even if one of the roots is zero – but if β = 0, say, then the complementary function will become $y = A e^{α x} + B$

CASE 2: $b^{2} - 4 a c = 0$ so that there is only one repeated root α

The complementary function is $y = (A + B x) e^{α x}$ where A and B are arbitrary constants

CASE 3: $b^{2} - 4 a c < 0$ so that α and β are complex conjugate roots that may be written as $p \pm q i$

The complementary function is $y = e^{p x} (A \cos q x + B \sin q x)$ where A and B are arbitrary constants
Note that if α and β are purely imaginary, then p = 0 and the complementary function becomes $y = A \cos q x + B \sin q x$

How do I solve a second order homogeneous differential equation?

STEPS 1 & 2: Use the auxiliary equation to find the differential equation’s complementary function (see above)

The complementary function, with its arbitrary constants A and B, is the general solution to the differential equation

STEP 3: If there are initial or boundary conditions associated with the differential equation, use these to find the values of the general solution’s arbitrary constants

This gives the particular solution to the differential equation with the given initial or boundary conditions
Note that because there are two arbitrary constants, you will require two separate initial or boundary conditions to find both constants’ values
If the initial or boundary conditions involve $\frac{d y}{d x}$ you will need to differentiate your general solution to find $\frac{d y}{d x}$ in terms of the constants A and B
Finding the values of A and B may require solving simultaneous equations

Worked example

Find the general solution to each of the following differential equations:

(i)

\frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 12 y = 0

(ii)

\frac{d^{2} y}{d x^{2}} + 6 \frac{d y}{d x} + 9 y = 0

(iii)

\frac{d^{2} y}{d x^{2}} - 4 \frac{d y}{d x} + 13 y = 0

8-2-1-aux-eqns--compl-functns-a-we-solution

For each of the differential equations in part (a), find the particular solution that satisfies

y = 1

and

\frac{d y}{d x} = 1

when

x = 0

8-2-1-aux-eqns--compl-functns-b-we-solution

Particular Integral

What is a particular integral?

A particular integral is part of the solution to a second order non-homogeneous differential equation of the form $a \frac{d^{2} y}{d x^{2}} + b \frac{d y}{d x} + c y = f (x)$
- When the particular integral is substituted into the left hand side it will produce f(x)
- The other part of the solution is the complementary function associated with the differential equation

How do I find the particular integral for a second order non-homogeneous differential equation?

STEP 1: Choose the correct ‘test form’ of particular integral, based on the function f(x) on the right-hand side of the non-homogeneous differential equation:

Form of f(x)	‘Test form’ of p.i	Notes
p	λ	p is a given constant λ is a constant to be found
p + qx	λ + μx	p & q are given constants λ & μ are constants to be found Use the full test form of the p.i., even if there is no constant term (i.e., even if p = 0)
p + qx + rx²	λ + μx + νx²	p, q & r are given constants λ, μ & ν are constants to be found Use the full test form of the p.i., even if there is no constant or x term (i.e., even if p = 0 and/or q =0)
pe^kx	λe^kx	k & p are given constants λ is a constant to be found
p cos ωx + q sin ωx	λ cos ωx + μ sin ωx	p, q & ω are given constants λ & μ are constants to be found Use the full test form of the p.i., even if f(x) only contains sin or only contains cos (i.e., even if p = 0 or q =0)

If all or part of the ‘test form’ of the particular integral occurs in the differential equation’s complementary function, you will need to modify the ‘test form’
See the following section for how to do this

STEP 2: Find the first and second derivatives of the ‘test form’ of particular integral
STEP 3: Substitute the first and second derivatives, along with the ‘test form’ itself, into the differential equation
STEP 4: By comparing coefficients, determine the correct values to use for the constants in the ‘test form’

This may require solving simultaneous equations involving the various constants
The ‘test form’ with the correct values of its constants is the particular integral for the differential equation

What if a part of the ‘test form’ of the particular integral already occurs in the differential equation’ complementary function?

Because the terms of the complementary function are solutions to the homogeneous equation $a \frac{d^{2} y}{d x^{2}} + b \frac{d y}{d x} + c y = 0$ , they cannot also provide possible solutions to the non-homogeneous equation $a \frac{d^{2} y}{d x^{2}} + b \frac{d y}{d x} + c y = f (x)$
If the standard ‘test form’ for the particular integral contains terms that already occur as a part of the complementary function, then the ‘test form’ needs to be modified by adding a factor of x (or possibly of powers of x) to the terms of the ‘test form’
Some examples:

The equation $\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} - 2 y = 3 e^{x}$ has complementary function $A e^{x} + B e^{- 2 x}$

The standard ‘test form’ of p.i. would be λe^x, but e^x times a constant already occurs in the complementary function
Therefore $λ x e^{x}$ would be used as the ‘test form’ instead

The equation $\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = 7$ has complementary function $A e^{- x} + B$

The standard ‘test form’ of p.i. would be λ, but a constant (B) already occurs in the complementary function
Therefore $λ x$ would be used as the ‘test form’ instead

The equation $\frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} + y = 5 e^{- x}$ has complementary function $(A + B x) e^{- x}$

The standard ‘test form’ of p.i. would be λe^-x, but e^-x times a constant AND xe^-x times a constant both already occur in the complementary function
Therefore $λ x^{2} e^{- x}$ would be used as the ‘test form’ instead

How do I solve a second order non-homogeneous differential equation?

STEP 1: Use the auxiliary equation to find the differential equation’s complementary function (‘c.f.’)
STEP 2: Find the differential equation’s particular integral (‘p.i.’) including the correct values of any constants
STEP 3: The general solution to the differential equation is the sum of the complementary function and the particular integral

I.e., the general solution is y = c.f. + p.i.

STEP 4: If there are initial or boundary conditions associated with the differential equation, use these to find the values of the general solution’s arbitrary constants

This gives the particular solution to the differential equation with the given initial or boundary conditions
Note that because there are two arbitrary constants (A and B from the complementary function), you will require two separate initial or boundary conditions to find both constants’ values
If the initial or boundary conditions involve $\frac{d y}{d x}$ you will need to differentiate your general solution to find $\frac{d y}{d x}$ in terms of the constants A and B
Finding the values of A and B may require solving simultaneous equations

Examiner Tip

Don’t forget to include the complementary function in your solution to a second order non-homogeneous differential equation – the solution is incomplete without it!
If your attempt to find the constants for a particular integral breaks down and doesn’t appear to have a solution, make sure that you have not used a ‘test form’ of the p.i. that includes terms also found in the complementary function
Finding the constants for the particular integral can be a very fiddly and algebra-heavy process – be sure to work slowly and methodically to avoid mistakes!

Worked example

Find the general solution to the differential equation

\frac{d^{2} y}{d x^{2}} + 5 \frac{d y}{d x} + 6 y = 12 x

8-2-1-particular-integral-a-we-solution

Find the general solution to the differential equation

\frac{d^{2} y}{d x^{2}} - 4 \frac{d y}{d x} + 3 y = 8 e^{3 x}

8-2-1-particular-integral-b-we-solution

Find the particular solution to the differential equation

\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} = 5 \sin x

that satisfies

y = 0

and

\frac{d y}{d x} = 1

when

x = 0

8-2-1-particular-integral-c-we-solution

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Solving Second Order Differential Equations (Edexcel A Level Further Maths: Core Pure)

Revision Note

What is a second order differential equation?

What are the types of second order differential equation?

How can I solve simple second order differential equations?

What is a complementary function?

What is an auxiliary equation?

How do I use the auxiliary equation to find the associated complementary function?

How do I solve a second order homogeneous differential equation?

What is a particular integral?

How do I find the particular integral for a second order non-homogeneous differential equation?

What if a part of the ‘test form’ of the particular integral already occurs in the differential equation’ complementary function?

How do I solve a second order non-homogeneous differential equation?

You've read 0 of your 10 free revision notes

Unlock more, it's free!

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Author: Roger