Basic Mathematics · Unit 10
Differential Equations
Exam-focused notes for Differential Equations (Basic Mathematics, MTH104): what the TU syllabus asks and how it has actually been tested, with 8 solved past questions from this unit.
What this unit covers
- First order differential equations
- Separable differential equations
- Linear differential equations
- Initial value problems
- Second order linear differential equations
- Homogeneous differential equations
- Phase lines and qualitative analysis
- Newton's method for root finding
Initial value problems
Question
What is initial value problem? Find the solution of the initial value problem $x\frac{dy}{dx} - y = x^2$, $y(2) = 5$. Evaluate: $\lim_{x \to \infty} \frac{3x^2 - 5x + 2}{5x^2 + 8x + 7}$. [1+4+5]
An Initial Value Problem (IVP) is a differential equation together with the value of the unknown function (and possibly its derivatives) specified at a single point, called the initial condition. General form: $$\frac{dy}{dx} = f(x,y), \qquad y(x0) = y0$$ T...
Full solved answer →Find the initial value problem in $\frac{dy}{dx} + 2y = 3$, $y(0) = 1$. [5]
- Differential equation: $\dfrac{dy}{dx} + 2y = 3$ - Initial condition: $y(0) = 1$ All required data is present. Identify form: First-order linear ODE $$\frac{dy}{dx} + P(x)y = Q(x), \quad P(x) = 2, \quad Q(x) = 3$$ Integrating factor: $$\mu(x) = e^{\int 2\...
Full solved answer →Newton's method for root finding
Use Newton's method to find $\sqrt[4]{10}$ correct to four decimal places. [5]
- Target value: $\sqrt[4]{10}$ (fourth root of 10) - Required accuracy: 4 decimal places Let $x = \sqrt[4]{10}$, so $x^4 = 10$, giving: $$f(x) = x^4 - 10 = 0$$ $$f'(x) = 4x^3$$ $$x{n+1} = xn - \frac{f(xn)}{f'(xn)} = xn - \frac{xn^4 - 10}{4xn^3} = \frac{3xn^...
Full solved answer →First order differential equations
What is a first order linear differential equation? Solve the initial value problem: $t\frac{dy}{dt} + 2y = t^3$, $t > 0$, $y(2) = 1$. [1+4]
A first order linear differential equation is one that can be written in the standard form: $$\frac{dy}{dt} + P(t)y = Q(t)$$ where $P(t)$ and $Q(t)$ are functions of the independent variable $t$ alone. It is called first order because it involves only the f...
Full solved answer →Linear differential equations
Solve the following differential equation: $x\frac{dy}{dx} = x^2 + 3y$, $x > 0$. [5]
- Differential equation: $x\dfrac{dy}{dx} = x^2 + 3y$ - Domain: $x 0$ Divide by $x$ (valid since $x 0$): $$\frac{dy}{dx} = x + \frac{3y}{x}$$ $$\frac{dy}{dx} - \frac{3}{x}y = x$$ This is linear: $\dfrac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = -\dfrac{3}{x}$, $...
Full solved answer →Separable differential equations
Solve: $xy' = y$, when $y(1) = 2$. [5]
- Differential equation: $xy' = y$ - Initial condition: $y(1) = 2$ All required data is present. Step 1: Separate variables $$x\frac{dy}{dx} = y$$ $$\frac{dy}{y} = \frac{dx}{x}$$ Step 2: Integrate both sides $$\int \frac{dy}{y} = \int \frac{dx}{x}$$ $$\lny ...
Full solved answer →Second order linear differential equations
Solve: $y'' - y' - 6y = 0$. [5]
Given data: - Differential equation: $y'' - y' - 6y = 0$ - Constant coefficients: $1, -1, -6$ - Homogeneous, second-order, linear ODE. All data present. Characteristic equation: Assume $y = e^{rx}$, so $y' = re^{rx}$, $y'' = r^2 e^{rx}$. Substituting: $$r^2...
Full solved answer →Phase lines and qualitative analysis
Draw a phase line for the equation $\frac{dy}{dx} = (y + 1)(y - 2)$ and use it to sketch solutions to the equation. [5]
- Autonomous ODE: $\dfrac{dy}{dx} = (y+1)(y-2)$ Set $\dfrac{dy}{dx} = 0$: $$(y+1)(y-2) = 0 \implies y = -1 \quad \text{or} \quad y = 2$$ These are the equilibrium solutions (constant solutions). Region Test Point $(y+1)$ $(y-2)$ $\frac{dy}{dx}$ Motion -----...
Full solved answer →Make Unit 10 stick
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