Further Mathematics | First order differential equations

👋 Welcome to First Order Differential Equations (FP2)

Hey mathematicians! This chapter, First Order Differential Equations, is where Pure Mathematics really connects with the real world. Differential equations (or DEs) are essentially the language used to describe change—how populations grow, how substances decay, or how temperature shifts over time.
Don't worry if this sounds intimidating! We will break down three powerful methods used exclusively in FP2 to solve the trickiest types of first order equations. Mastering these techniques will significantly boost your problem-solving toolkit!

Prerequisite Check: Before diving in, make sure you are comfortable with basic integration, the Product Rule for differentiation, and the chain rule.

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Section 1: Understanding First Order Equations

A First Order Differential Equation involves only the first derivative, \( \frac{dy}{dx} \). We are looking for the function \( y = f(x) \) that satisfies the equation.
In FP2, we focus on methods beyond simple separation of variables.

Type 1: Homogeneous Equations

A first order equation is homogeneous if it can be written in the form:

$$ \frac{dy}{dx} = f\left(\frac{y}{x}\right) $$

What does that mean? If you look at the degrees of the terms in the numerator and the denominator, they are the same (or can be made the same). If you substitute \(\lambda x\) for \(x\) and \(\lambda y\) for \(y\), the \(\lambda\)s cancel out.

Example: \( \frac{dy}{dx} = \frac{x^2 + y^2}{xy} \). Every term (\(x^2\), \(y^2\), \(xy\)) has a total power of 2. It is homogeneous!

The Homogeneous Solution Method: \( y = vx \)

The trick to solving a homogeneous equation is the substitution:

$$ \mathbf{y = vx} $$

where \(v\) is a function of \(x\). This substitution magically transforms the problem into one you can solve using separation of variables.

Step-by-Step Process:

1. Substitute \(y\): Replace \(y\) with \(vx\). (This is what makes the right side simplify to a function of \(v\) only).
2. Find \(\frac{dy}{dx}\): Differentiate \(y = vx\) using the Product Rule.

   $$ \frac{dy}{dx} = v \cdot (1) + x \cdot \frac{dv}{dx} $$ $$ \frac{dy}{dx} = v + x \frac{dv}{dx} $$

3. Substitute into the DE: Replace \(\frac{dy}{dx}\) in the original equation.
4. Separate Variables: The resulting equation will always be separable in terms of \(v\) and \(x\). Isolate \(\frac{dv}{dx}\) and rearrange so that all \(v\) terms are with \(dv\) and all \(x\) terms are with \(dx\).
5. Integrate and Substitute Back: Integrate both sides and finally replace \(v\) with \( \frac{y}{x} \) to find the general solution for \(y\).

🧠 Memory Aid: When you see \(y/x\) repeated in the equation, think Very Xtra work is needed! (\(y=vx\)).

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Section 2: Linear First Order Equations (The Integrating Factor)

The most common and important type of first-order DE you will encounter in FP2 is the Linear First Order Equation.

The Standard Linear Form

An equation is linear if it can be written in the standard form:

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

Here, \(P(x)\) and \(Q(x)\) are functions of \(x\) only (or constants). Note that the coefficient of \(\frac{dy}{dx}\) must be 1. If it isn't, divide the entire equation by that coefficient first!

The Integrating Factor (IF)

We use a special function called the Integrating Factor, \(I(x)\), to solve these equations. The IF makes the left side of the equation immediately integrable by turning it into the result of the Product Rule.

The formula for the Integrating Factor is:

$$ \mathbf{I(x) = e^{\int P(x) dx}} $$

IMPORTANT: The \(P(x)\) must be the coefficient of \(y\) in the standard form.

The Solution Method: LIFe (Linear Integrating Factor)

This method is a systematic process. Follow these steps precisely:

1. Standard Form Check: Ensure the equation is in the form \( \frac{dy}{dx} + P(x)y = Q(x) \). Identify \(P(x)\).
2. Calculate the IF: Find \( I(x) = e^{\int P(x) dx} \). (Crucially, you do not need the \(+C\) at this stage).
3. Multiply: Multiply the entire standard equation by the IF, \(I(x)\).
4. The Product Rule Shortcut: The entire left side is now guaranteed to be the derivative of the product \( y \cdot I(x) \):

   $$ \mathbf{\frac{d}{dx} \left(y \cdot I(x)\right) = Q(x) \cdot I(x)} $$

5. Integrate: Integrate both sides with respect to \(x\):

   $$ y \cdot I(x) = \int Q(x) I(x) dx + C $$

6. Isolate \(y\): Divide by \(I(x)\) to get the final solution for \(y\).

💡 Why does this work? (Did you know?)
The IF is designed specifically so that when you multiply it by \(P(x)y\), the resulting expression \(I(x) P(x) y\) provides the missing term needed to complete the Product Rule derivative of \(y \cdot I(x)\). It’s a mathematical trick to simplify integration!

⚠️ Common Mistakes with the IF Method

• Sign Error: If the standard form is \( \frac{dy}{dx} - 3y = x \), then \(P(x)\) is \(-3\), not just 3. The sign is crucial for calculating \(I(x)\).
• Missing Constant: Forgetting the \(+C\) after the final integration step (Step 5). This changes the general solution to a specific solution!
• Not Standard Form: Starting the method when the coefficient of \(\frac{dy}{dx}\) is not 1. (e.g., in \(x \frac{dy}{dx} + y = 2x^2\), you must first divide by \(x\)).

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Section 3: Equations Reducible to Linear Form (Bernoulli's Equation)

Sometimes, a DE looks almost linear, but a pesky term prevents us from using the Integrating Factor directly. These are often known as Bernoulli's Equations (though the name isn't required, recognizing the form is!).

The Reducible Form

The equation looks like this:

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

It's linear, except for the extra power of \(y\) on the right-hand side (\(y^n\), where \(n \neq 0\) and \(n \neq 1\)).

The Key Substitution

To solve this, we must transform the equation into a standard linear form using a substitution involving a new variable, \(z\).

$$ \mathbf{z = y^{1-n}} $$

Don't panic! The process is formulaic, but requires careful algebra and application of the Chain Rule.

The Step-by-Step Transformation

1. Divide: Divide the entire equation by \(y^n\):

   $$ y^{-n} \frac{dy}{dx} + P(x)y^{1-n} = Q(x) $$

2. The Substitution: Let \( z = y^{1-n} \).
3. Find \(\frac{dz}{dx}\): Use the Chain Rule to find the derivative of \(z\) with respect to \(x\):

   $$ \frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx} $$ $$ \frac{dz}{dx} = (1-n) y^{1-n-1} \frac{dy}{dx} $$ $$ \frac{dz}{dx} = (1-n) y^{-n} \frac{dy}{dx} $$

4. Transform the \(\frac{dy}{dx}\) term: From Step 3, we can see that:

   $$ y^{-n} \frac{dy}{dx} = \frac{1}{1-n} \frac{dz}{dx} $$

5. Rewrite the Equation: Substitute the new terms back into the divided equation (Step 1).

   $$ \frac{1}{1-n} \frac{dz}{dx} + P(x)z = Q(x) $$

6. Standard Linear Form (in \(z\)): Multiply by \((1-n)\) to get the standard form:

   $$ \mathbf{\frac{dz}{dx} + (1-n)P(x)z = (1-n)Q(x)} $$

Congratulations! You now have a standard linear DE in the variable \(z\). You can now solve this using the Integrating Factor method from Section 2.

Final Step: Once you have solved for \(z(x)\), remember to replace \(z\) with \(y^{1-n}\) to find the final solution for \(y\).

General Strategy Tip

Always look at the structure of the equation first:
1. Does it have \(\frac{dy}{dx}\) and \(y\) (Linear)? Use IF.
2. Does it have \(\frac{dy}{dx}\), \(y\), AND a multiplicative \(y^n\) term (Reducible)? Use \(z = y^{1-n}\).
3. Can it be written as a function of \(y/x\) (Homogeneous)? Use \(y = vx\).