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[h] IB Mathematics AA HL Flashcards -differential equations
[q] Differential equations
If we take \( y \) as a function of \( x \) (i.e., \( y = f(x) \)), then a **differential equation** is an equation that relates \( x, y \) and \( \frac{dy}{dx} \) (or higher derivatives), with general form:
\[
F(x, y, \frac{dy}{dx}, \frac{d^2y}{dx^2}, \ldots, \frac{d^ky}{dx^k}) = 0
\]
where \( k \) is the order of the differential equation.
E.G. :
\[
\frac{dy}{dx} + y = x – 2
\]
These are three ways of writing the same differential equation. It means, in words: The derivative of a function \( f'(x) \) plus that function of \( x \) equals \( x – 2 \).
From this example, you start to see that solving a differential equation is not a case of finding a number, but rather finding the unknown function \( y = f(x) \) that fits. In IB calculus, you will only deal with first-order differential equations.
[q]
SEPARABLE EQUATIONS
This is the first of a few main types of differential equations that we will see & solve. It may also be the simplest of our types to solve, essentially just combining a couple of skills you already know.
[a]
FORM: A separable equation is one that is in the form:
\[
\frac{dy}{dx} = P(x) – Q(y) \quad \text{… or it can at least be rearranged into this form}
\]
[q]
SOLVING: To find the unknown function of \( x \), \( y = f(x) \), follow these steps:
1. Write in the form: \( \frac{dy}{dx} = P(x)Q(y) \)
2. Rearrange to get \( Q(y) dy = P(x) dx \)
3. Integrate both sides to get \( \int Q(y) dy = \int P(x) dx \)
4. Solve for \( y \)
[a]
1st ORDER LINEAR DIFFERENTIAL EQUATIONS
As well as being first order, if the equation also just has the first power of \( y \) only, it is called linear. Just as with separable equations, there are many forms these could be written in, but there is a standard form:
\[
\frac{dy}{dx} + P(x)y = Q(x)
\]
[q]
INTEGRATING FACTOR: Looking at the standard form above, the left-hand side (LHS) is close to the result that one might get after using the product rule. We can use this to our advantage when trying to find solutions to these types of equations:
\[
\frac{dy}{dx} + P(x)y = Q(x) \quad \text{multiply by } I(x), \text{ a function:}
\]
If \( I(x) = e^{\int P(x) dx} \), then the LHS is the result of the product rule:
\[
\frac{d}{dx}(I(x)y) = I(x)Q(x)
\]
[a]
So we have a solution, as long as we know what this \( I(x) \) function is. We know that:
\[
I(x) = e^{\int P(x)dx}
\]
E.G. : Solve the differential equation \( y’ – 3y = e^x \):
1. In the correct form ✔
2. Calculate integrating factor: \( I(x) = e^{\int -3dx} = e^{-3x} \)
3. Multiplying by \( I(x) \):
\[
e^{-3x}y’ – 3y e^{-3x} = e^{-3x} e^x
\]
4. Integrate both sides [product rule on LHS]:
\[
y e^{-3x} = \int e^{-2x} dx = -\frac{1}{2} e^{-2x} + C
\]
5. Solve for \( y \):
\[
y = -\frac{1}{2} e^{x} + Ce^{3x}
\]
[q]
HOMOGENEOUS DIFFERENTIAL EQUATIONS
In general, a homogeneous function is a function of \( x \) and \( y \) such that each term’s powers of \( x \) and \( y \) add up to \( n \) (for example: \( g(x,y) = 4x^3y^{-2} + y^3 + 0.1y^5 \) powers sum to 5 in every term). An extension of this is the homogeneous differential equation, which is when we have:
\[
\frac{dy}{dx} = \frac{F(x,y)}{G(x,y)}
\]
with \( F(x,y) \) and \( G(x,y) \) both being homogeneous of the same degrees \( n \). Also, by dividing by \( x^n \), it can be written in the form:
\[
\frac{dy}{dx} = F\left(\frac{y}{x}\right)
\]
From this, we will discover a method for solving them, just like the previous types of differential equations.
[a]
METHOD for solving 1st order linear differential equations is therefore:
1. Rearrange to \( \frac{dy}{dx} = \frac{F(x,y)}{G(x,y)} \), then divide by \( x^n \) to give \( \frac{dy}{dx} = F\left(\frac{y}{x}\right) \)
2. Let \( y = vx \). By product rule, \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substitute \( \frac{dy}{dx} \) out for \( v \) and \( v + x\frac{dv}{dx} \) respectively.
3. Solve, using the separable equations method.
4. Substitute \( v \) back for \( y \). Find \( c \) if possible.
E.G. 10: Solve \( \frac{dy}{dx} = \frac{3x^2 + y^2}{xy} \), when \( y(1) = 1 \):
1. Already in the form \( \frac{F(x,y)}{G
E.G. : Solve \( \frac{dy}{dx} = \frac{3x^2 + y^2}{xy} \), when \( y(1) = 1 \):
1. Already in the form \( \frac{F(x,y)}{G(x,y)} \). Degree is 2, so we divide by \( x^2 \):
\[
\frac{dy}{dx} = \frac{3 + \left(\frac{y}{x}\right)^2}{\frac{y}{x}} \Rightarrow v = \frac{y}{x}
\]
2. Substitute \( y = vx \):
\[
\frac{dy}{dx} = v + x \frac{dv}{dx}
\]
3. Substitute into the equation:
\[
v + x \frac{dv}{dx} = \frac{3 + v^2}{v}
\]
4. Rearranging gives:
\[
x \frac{dv}{dx} = \frac{3}{v} – v – 1
\]
[q]
EULER’S METHOD
This section won’t deal with a specific type of differential equation, nor will we be finding general solutions for the function \( y \). Instead, we will see a method for approximating a \( y \)-value, given an \( x \)-value and a set of initial coordinates. As you know an expression for \( \frac{dy}{dx} \), you can find the equation of the tangent passing through this starting point. You could then move along that tangent until your final \( x \)-value, but as you can see below, it would be very inaccurate.
[a]
If we could break up the \( x \)-difference into small, regular intervals and correct the slope at each point, it will be far more accurate. This method of approximating \( y \)-values is called Euler’s Method.
FORMAL METHOD: To sum up the process above, where we find each \( y \)-value and each slope, then the next point, etc., to find our final \( y \)-value:
\[
\text{For differential equation } \frac{dy}{dx} = F(x,y) \text{ and initial condition } y(x_0) = y_0
\]
If you know the \( n^{th} \) point \( (x_n, y_n) \) and step size \( h \):
\[
(n+1)^{th} \text{ point is given by } x_{n+1} = x_n + h \text{ and } y_{n+1} = y_n + h F(x_n, y_n)
\]
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