8.1.2 Solving First Order Differential Equations

General Solution of First-Order Differential Equations

A first-order differential equation has the form:

\frac{dy}{dx} = f(x, y)

Using substitution, we rewrite the equation in terms of new variables to simplify its form.

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Example Substitution: $z = x + y$

Given a differential equation:

\frac{dy}{dx} = \frac{x + y + c_1}{x + y + c_2}

the substitution $z = x + y$ simplifies the equation because it eliminates $x + y$ as a combined term.

Introduction to Substitution in Differential Equations

Substitution is a powerful technique for simplifying differential equations, making them easier to solve. It is especially useful when the equation contains combinations of variables that suggest a change of variables.

Steps for Solving Using Substitution

Identify an appropriate substitution $z = g(x, y)$
Differentiate $z$ with respect to $x$ , and express $\frac{dy}{dx}$ in terms of $z$ and $\frac{dz}{dx}$
Substitute into the original differential equation to rewrite it in terms of $z$ and $\frac{dz}{dx}$
Solve the resulting equation for $z$
Back-substitute $z = g(x, y)$ to find the solution for $y$ .

Finding Tangents to Polar Curves

For differential equations in polar coordinates $r$ and $\theta$ , the slope of a tangent at any point is given by:

\frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta}

Using this formula, we can find tangents that are:

Parallel to the initial line ( $y = 0$ ), where $\frac{dy}{dx} = 0$ , or:

r' \sin \theta + r \cos \theta = 0

Perpendicular to the initial line ( $x = 0$ ), where $\frac{dy}{dx} \to \infty$ , or:

r' \cos \theta - r \sin \theta = 0

The Integrating Factor for F.O.D.E.s

An integrating factor can simplify equations of the form:

\frac{dy}{dx} + P(x)y = Q(x)

Multiplying through by the integrating factor $F(x)$ , where:

F(x) = e^{\int P(x) \, dx}

transforms the equation into a product rule:

F(x) \frac{dy}{dx} + F(x) P(x) y = F(x) Q(x) \implies \frac{d}{dx}[F(x)y] = F(x)Q(x)

This allows for direct integration to solve for $y$ .

Worked Examples

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Example 1: Substitution in a Differential Equation

Solve $\frac{dy}{dx} = \frac{x + y + 3}{x + y - 1}$ using the substitution $z = x + y$

Step 1: Differentiate the Substitution

z = x + y \implies \frac{dz}{dx} = 1 + \frac{dy}{dx}

Rearrange:

\frac{dy}{dx} = \frac{dz}{dx} - 1

Step 2: Substitute into the Equation

Substitute $\frac{dy}{dx} = \frac{dz}{dx} - 1$ and $z = x + y$ into the original equation:

\frac{dz}{dx} - 1 = \frac{z + 3}{z - 1}

Simplify:

\frac{dz}{dx} = \frac{z + 3}{z - 1} + 1 = \frac{z + 3 + z - 1}{z - 1} = \frac{2z + 2}{z - 1}

Factorise:

\frac{dz}{dx} = \frac{2(z + 1)}{z - 1}

Step 3: Solve the Resulting Equation

Rearrange for integration:

\int \frac{z - 1}{z + 1} \, dz = \int 2 \, dx

Expand $\frac{z - 1}{z + 1}$

\frac{z - 1}{z + 1} = 1 - \frac{2}{z + 1}

Integrate:

\int (1 - \frac{2}{z + 1}) \, dz = \int 2 \, dx

z - 2 \ln |z + 1| = 2x + c

Step 4: Back-Substitute $z = x + y$

Substitute $z = x + y$

x + y - 2 \ln |x + y + 1| = 2x + c

Simplify:

y - 2 \ln |x + y + 1| = x + c

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Example 2: Tangents Parallel to and Perpendicular to the Initial Line

Find tangents parallel and perpendicular to the initial line for the polar curve $r = 1 + \cos \theta$

Step 1: Find $r′$

Differentiate $r$ with respect to $\theta$

r' = -\sin \theta

Step 2: Tangents Parallel to the Initial Line ( $y = 0$ )

Set:

r' \sin \theta + r \cos \theta = 0

Substitute $r = 1 + \cos \theta$

-\sin \theta \times \sin \theta + (1 + \cos \theta) \cos \theta = 0.

Simplify:

-\sin^2 \theta + \cos \theta + \cos^2 \theta = 0

\implies 1 - \sin^2 \theta + \cos \theta = 0

\implies 1 + \cos \theta = \sin^2 \theta

Solve for $\theta$ within $[0, 2\pi]$

Step 3: Tangents Perpendicular to the Initial Line ( $x = 0$ )

Set:

r' \cos \theta - r \sin \theta = 0

Follow similar steps to determine $\theta$ .

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Example 3: Using the Integrating Factor

Solve:

\frac{dy}{dx} + \frac{3}{x}y = \frac{\sin x}{x^3}

Step 1: Identify $P(x)$

Here, $P(x) = \frac{3}{x}$

Step 2: Find the Integrating Factor

F(x) = e^{\int P(x) \, dx} = e^{\int \frac{3}{x} \, dx} = e^{3 \ln x} = x^3

Step 3: Multiply Through by $F(x)$

x^3 \frac{dy}{dx} + 3x^2 y = \sin x

Rewrite as:

\frac{d}{dx}(x^3 y) = \sin x

Step 4: Integrate

x^3 y = \int \sin x \, dx = -\cos x + c

Solve for $y$ :

y = \frac{-\cos x + c}{x^3}

Note Summary

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Common Mistakes

Forgetting to differentiate substitutions correctly (e.g., $z = x + y$ )
Skipping steps when rearranging differential equations.
Neglecting to back-substitute the original variables after solving.
Misapplying the integrating factor $F(x)$ by omitting $e^{\int P(x) \, dx}$
Incorrectly evaluating boundary conditions or constants of integration.

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Key Formulas

Substitution Rule:

z = g(x, y) \quad \implies \quad \frac{dz}{dx} = \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}\frac{dy}{dx}

Area in Polar Coordinates:

A = \frac{1}{2} \int_\alpha^\beta r^2 \, d\theta

Slope of Tangent in Polar Coordinates:

\frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta}

Integrating Factor:

F(x) = e^{\int P(x) \, dx}

Solving First Order Differential Equations Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

8.1.2 Solving First Order Differential Equations

General Solution of First-Order Differential Equations

Example Substitution: $z = x + y$

Introduction to Substitution in Differential Equations

Steps for Solving Using Substitution

Finding Tangents to Polar Curves

The Integrating Factor for F.O.D.E.s

Worked Examples

Example 1: Substitution in a Differential Equation

Example 2: Tangents Parallel to and Perpendicular to the Initial Line

Example 3: Using the Integrating Factor

Note Summary

Common Mistakes

Key Formulas

500K+ Students Use These Powerful Tools to Master Solving First Order Differential Equations For their A-Level Exams.

Other Revision Notes related to Solving First Order Differential Equations you should explore

Solving First Order Differential Equations Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

8.1.2 Solving First Order Differential Equations

General Solution of First-Order Differential Equations

Example Substitution: z=x+yz = x + yz=x+y

Introduction to Substitution in Differential Equations

Steps for Solving Using Substitution

Finding Tangents to Polar Curves

The Integrating Factor for F.O.D.E.s

Worked Examples

Example 1: Substitution in a Differential Equation

Example 2: Tangents Parallel to and Perpendicular to the Initial Line

Example 3: Using the Integrating Factor

Note Summary

Common Mistakes

Key Formulas

500K+ Students Use These Powerful Tools to Master Solving First Order Differential Equations For their A-Level Exams.

Other Revision Notes related to Solving First Order Differential Equations you should explore

Example Substitution: $z = x + y$