Differential Equations



		Differential Equations What is a differential equation? A differential equation contains one or more terms involving derivatives of one variable (the dependent variable, y) with respect to another variable (the independent variable, x). For example, What does the solutions of a differential equation look like? Unlike algebraic equations, the solutions of differential equations are functions and not just numbers. What exactly does a differential equation represent? It represents the relationship between a continuously varying quantity and its rate of change. This is very essential in all scientific investigation. Where are differential equations used in real life? In physics, chemistry, biology and other areas of natural science, as well as areas such as engineering and economics. This is a picture of wind engineering. What is an ordinary differential equation? A differential equation that involves a function of a single variable and some of its derivatives. For example, What is the order of a differential equation? The order of a differential equation is the order of the highest derivative that appears in the equation. The above examples are both first order differential equations. An example of a second order differential equation is . Method 1: Separation of Variables Do you know when to use this method? The separation of variables method is used when a differential equation can be written in the form $\frac{dy}{dx} = f(y)g(x)$ where f is a function of y only and g is a function of x only. What is an initial value problem (IVP)? An initial value problem is one in which some initial conditions are given to solve a differential equation. How do we solve problems using this method? Rewrite the problem as $\frac{1}{f(y)}dy = g(x)dx$ and integrate both sides. Use the initial condition to find the constant of integration. For example: Solve $\frac{dy}{dx} = xy$ with initial condition $y(0) = 1.$ Step 1: Divide through by y. We get $\frac{1}{y}\, \frac{dy}{dx} = x.$ Step 2: Integrate both sides with respect to x. We get $\intop \frac{1}{y}\, \frac{dy}{dx}dx = \intop x\, dx$ $\int \frac{1}{y}\, dy = \int x\, dx$ $\ln(y) = x + c$ Step 3: Make y the subject of the equation. $y = Ae^x$ where $A = e^c$ This is the general solution of the differential equation. Step 4: The initial condition $y(0) = 1$ means when $x = 0, y = 1$ . Substituting these values in the general solution gives A = 1. Hence the solution to the initial value problem is $y = e^x$ . Test Yourself Method 2: Integrating Factor When do we use the integrating factor method? The integrating factor method is used when the differential equation is (or can be rearranged) in the form $\frac{dy}{dx} + p(x)\,y = q(x)$ where p and q are functions of x only. How do we find the integrating factor? The integrating factor is $e^{\int p(x)\, dx}$ . How do we solve differential equations using this method? Rearrange the differential equation (if needed) to the standard form and find the integrating factor. Multiply through by the integrating factor and rewrite the left hand side with derivatives y $y e^{\int p(x)\, dx}$ . Integrate both sides gives the general solution. For example: Solve $\frac{dy}{dx} + y = e^{-x}$ with y(0) = 1. Step1: Compare the equation with the standard form $\frac{dy}{dx} + p(x)y = q(x)$ gives $p(x) = 1$ and $q(x) = e^{-x}$ . Step2: Find the integrating factor $e^{\int 1 \,dx} = e^x$ Step3: Multiply through by the integrating factor, we get $e^x \,\frac{dy}{dx} + y\,e^x = 1$ . Step4: Replace $e^x \,\frac{dy}{dx} + ye^x$ with derivative of $ye^x$ i.e. $\frac{d}{dt} \Big [ ye^x \Big ]$ . Step5: Integrate both sides and get $ye^x = x + c$ . Step6: Use the initial condition to find c. $x = 0$ , $y = 1$ gives $c = 1$ . Hence the solution to the problem is $ye^x = x + 1$ which is $y = (x + 1)e^{-x}$ Test Yourself




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