Tutorial Week 8

Initial Value Problems

With initial value problems, you’re given a set of initial conditions (i.e. \(y(a) = y_a\), \(y'(a) = y_b\)).

Q1: Solve the initial value problem with \(\frac{d^2y}{dx^2} = x + 2 + \cos x\), with \(y(0) = 1\) and \(y'(0) = 2\).

Separable Differential Equations

Separable differential equations are in the form of \(\frac{dy}{dx} = f(x) \cdot g(y)\), where f is a function in terms of x and g is a function in terms of y.

This differential equation can then be rewritten as \(\frac{1}{g(y)} dy = f(x) dx\), and then be integrated on both sides, \(\int \frac{1}{g(y)} dy = \int f(x) dx\).

After computing this integral, you’ll just need to solve for the variable y.

Q2: Solve the differential equation \(\frac{dy}{dx} = 3x^2y - 3x^2 - y + 1\).

Linear Differential Equations

A first-order linear differential equation the standard form is given by \(\frac{dy}{dx} + P(x)y = Q(x)\).

To solve a first-order linear differential equation, we define the integrating factor as \(I(x) = e^{\int P(x) dx}\), and then multiply both sides of \(\frac{dy}{dx} + P(x)y = Q(x)\) by \(I(x)\) to get \(I(x)(frac{dy}{dx} + P(x)y) = I(x)(Q(x))\).

With the choice of \(I(x)\), \(I(x)(\frac{dy}{dx} + P(x)y) = I(x)(Q(x))\) simplifies to \((I(x)y)' = I(x)Q(x)\).

We then integrate both sides (\(\int (I(x)y)' = \int I(x)Q(x) dx\)), which simplifies to \(I(x)y = \int I(x)Q(x) dx\), which we can use to solve for y.