Solving Equations Revision Notes for Edexcel A-Level Mathematics

10.2.1 Numerical Methods in Context

Non-Homogeneous Second-Order Differential Equation (S.O.D.E)

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Consider a second-order differential equation of the form:

a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = F(x)

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Example: Solve the differential equation:

\frac{d^2y}{dx^2} - 5 \frac{dy}{dx} + 6y = 13 \sin(3x)

Solution:

The only way to solve this equation is to assume that y is of the form $\lambda \sin(3x) + \mu \cos(3x)$ . Let's try this as a solution:

Assume the particular solution:

y = \lambda \sin(3x) + \mu \cos(3x)

Find the first and second derivatives:

\frac{dy}{dx} = 3\lambda \cos(3x) - 3\mu \sin(3x)

\frac{d^2y}{dx^2} = -9\lambda \sin(3x) - 9\mu \cos(3x)

Substitute into the original equation:

-9\lambda \sin(3x) - 9\mu \cos(3x) - 5[3\lambda \cos(3x) - 3\mu \sin(3x)] + 6[\lambda \sin(3x) + \mu \cos(3x)] = 13 \sin(3x)

Simplify the equation:

(-9\lambda + 15\mu + 6\lambda) \sin(3x) + (-9\mu - 15\lambda + 6\mu) \cos(3x) = 13 \sin(3x)

(15\mu - 3\lambda) \sin(3x) + (-15\lambda - 3\mu) \cos(3x) = 13 \sin(3x)

Match the coefficients of $\sin(3x)$ and $\cos(3x)$ with the corresponding terms on the right-hand side:

$15\mu - 3\lambda = 13$
$-15\lambda - 3\mu = 0$

Solve the system of equations:

15\mu - 3\lambda = 13

-15\lambda - 3\mu = 0

From the first equation:

\mu = \frac{1}{5}, \quad \lambda = -\frac{1}{6}

Therefore, the particular solution is:

y = -\frac{1}{6} \sin(3x) + \frac{5}{6} \cos(3x)

General Solution:

If $y = -\frac{1}{6} \sin(3x) + \frac{5}{6} \cos(3x)$ is a particular solution, then the general solution is:

y = -\frac{1}{6} \sin(3x) + \frac{5}{6} \cos(3x) + 0

The zero solutions are found by solving the associated homogeneous equation previously introduced.

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Solving the Homogeneous Equation

Given the differential equation:

\frac{d^2y}{dx^2} - 5 \frac{dy}{dx} + 6y = 0

Solve the characteristic equation: The characteristic equation is derived by assuming a solution of the form $y = e^{mx}$ , leading to:

m^2 - 5m + 6 = 0

Factorizing the quadratic:

(m - 2)(m - 3) = 0

This gives the roots m = 2 and m = 3.

Write the general solution to the homogeneous equation: The general solution (complementary function) is formed by the linear combination of the solutions corresponding to the roots:

y_h = Ae^{2x} + Be^{3x}

where $A$ and $B$ are arbitrary constants.

Combine with the particular integral: From the previous work, the particular integral (the particular solution to the non-homogeneous equation) was found to be:

y_p = -\frac{1}{6} \sin(3x) + \frac{5}{6} \cos(3x)

Form the complete solution: The complete solution to the non-homogeneous differential equation is the sum of the particular integral and the complementary function:

y = -\frac{1}{6} \sin(3x) + \frac{5}{6} \cos(3x) + Ac^{2x} + Bc^{3x}

Note on the complementary function: The complementary function on its own is not a solution to the original non-homogeneous equation (since it equals zero), but it complements the particular integral to form the full solution.

Problem Statement

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Solve the differential equation:

\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 3x^2

1. Complementary Function (C.F.)

First, solve the homogeneous part of the equation:

\text{C.F.:} \quad m^2 - 5m + 6 = 0

Factorizing:

(m - 2)(m - 3) = 0

This gives the roots m = 2 and m = 3.

Thus, the complementary function is:

y_h = Ae^{2x} + Be^{3x}

2. Particular Integral (P.I.)

Next, propose a form for the particular integral. Since the non-homogeneous term is $3x^2$ , assume:

y = ax^2 + bx + c

Then:

\frac{dy}{dx} = 2ax + b

\frac{d^2y}{dx^2} = 2a

Substitute $y$ and its derivatives into the differential equation:

2a - 5(2ax + b) + 6(ax^2 + bx + c) = 3x^2

Simplifying, collect terms involving $x^2, x,$ and constants:

6ax^2 + (6b - 10a)x + (2a - 5b + 6c) = 3x^2

Equating coefficients of like terms:

Coefficient of $x^2: 6a = 3 ⟹ a = \frac{1}{2}$
Coefficient of $x: 6b - 10\left(\frac{1}{2}\right) = 0 ⟹ b = \frac{5}{6}$
Constant term: $2\left(\frac{1}{2}\right) - 5\left(\frac{5}{6}\right) + 6c = 0 ⟹ c = \frac{19}{36}$ So, the particular integral is:

y_p = \frac{1}{2}x^2 + \frac{5}{6}x + \frac{19}{36}

3. General Solution

The general solution is the sum of the complementary function and the particular integral:

\boxed {y = \frac{1}{2}x^2 + \frac{5}{6}x + \frac{19}{36} + Ae^{2x} + Be^{3x}}

Forms of Particular Integrals

RHS	P.I.
k	c
$\lambda x+ \mu$	$bx+c<br>$ Note: $\mu$ can = 0
$\lambda x^2+ \mu x+2$	$ax^2+bx+c$ Note: $\mu$ , 2 can = 0
$\lambda e^{kx}$	$ae^{kx}$
$p\sin \lambda x+q\cos \lambda x$	$a\sin \lambda x+b\cos \lambda x$ Note: $a$ , $b$ can = 0

Anything more complemented will be given in the question.

S.O.D.E. Where P.I. is Part of C.F.

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Example Problem: Solve $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = e^{2x}$ .

Find the Complementary Function (C.F.):

First, solve the characteristic equation associated with the homogeneous part:

m^2 - 5m + 6 = 0

Factorize the characteristic equation:

(m - 2)(m - 3) = 0

This gives m = 2 and m = 3, so the C.F. is:

y = Ae^{2x} + Be^{3x}

Propose a Particular Integral (P.I.):

Since $e^{2x}$ is already part of the C.F., assume:

y = \lambda x e^{2x}

Differentiate this proposed solution:

\frac{dy}{dx} = \lambda x \times 2e^{2x}+e^{2x} \times \lambda =\lambda e^{2x} + 2\lambda x e^{2x}

\frac{d^2y}{dx^2} = 2\lambda e^{2x} + 2\lambda e^{2x}+ 4\lambda x e^{2x} = 4\lambda e^{2x} + 4\lambda x e^{2x}

Substitute into the differential equation:

Substitute $y = \lambda x e^{2x}, \frac{dy}{dx}$ , and $\frac{d^2y}{dx^2}$ into the original equation:

\text{LHS} = 4\lambda e^{2x} + 4\lambda x e^{2x} - 5(\lambda e^{2x} + 2\lambda x e^{2x}) + 6(\lambda x e^{2x})

Simplify the equation:

e^{2x}(4\lambda + 4\lambda x - 5\lambda - 10\lambda x + 6\lambda x) = e^{2x}

e^{2x}(-\lambda) = e^{2x}

\lambda = -1

Final Solution:

The Particular Integral is:

y = -x e^{2x}

Therefore, the general solution is:

y = (A - x)e^{2x} + Be^{3x}

Justification of Trying $\lambda x e^{2x}$ :

Start with the given S.O.D.E.:

\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = e^{2x}

Rewrite the differential equation:

This can be factored as:

\left(\frac{d}{dx} - 2\right)\left(\frac{d}{dx} - 3\right)y = e^{2x}

Multiply both sides by $\left(\frac{d}{dx} - 2\right)$ to get:

\left(\frac{d}{dx} - 2\right)^2\left(\frac{d}{dx} - 3\right)y = \left(\frac{d}{dx} - 2\right)e^{2x}

Calculate the right-hand side:

Simplify the expression on the right:

\left(\frac{d}{dx} - 2\right)e^{2x} = \frac{d}{dx} e^{2x} - 2e^{2x} = 2e^{2x} - 2e^{2x} = 0

This results in:

\left(\frac{d}{dx} - 2\right)^2\left(\frac{d}{dx} - 3\right)y = 0

Conclusion:

This leads to the equation:

\left(\frac{d}{dx} - 2\right)^2\left(\frac{d}{dx} - 3\right)y = 0

The repeated root at m = 2 results in the solution of the form:

y = (A + Bx)e^{2x}

Here, the term $Bx e^{2x}$ is equivalent to the form $\lambda x e^{2x}$ that was tried, justifying its use as a trial solution.

Solving Equations (Edexcel A-Level Mathematics): Revision Notes

10.2.1 Numerical Methods in Context

Non-Homogeneous Second-Order Differential Equation (S.O.D.E)

Solution:

Solving the Homogeneous Equation

Problem Statement

1. Complementary Function (C.F.)

2. Particular Integral (P.I.)

3. General Solution

Forms of Particular Integrals

S.O.D.E. Where P.I. is Part of C.F.

Example Problem: Solve $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = e^{2x}$ .

Justification of Trying $\lambda x e^{2x}$ :

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