Differential equation-Lecture

Differential Equations Instructor by Asst.Prof.Siriwan Wasukree

Contents 1 Definitions and Concepts 2 First-order Ordinary Differential Equation 3 Separation of Variables 4 Homogeneous Equations 5 Exact Equations 6 Linear Equations 7 Applications by Aj.Weerawat

1 Definitions and Concepts A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives. Here are a few more examples of differential equations. (1) (2) (3) (4) (5) (6)

Order The order of a differential equation is the largest derivative present in the differential equation. In the differential equations listed above (1) is a first order differential equation, (2), (4) and (5) are second order differential equations, (6) is a third order differential equation and (3) is a fourth order differential equation.

Solution A solution to a differential equation on an interval a < t < b is any function y(t) which satisfies the differential equation in question on the interval a < t < b It is important to note that solutions are often accompanied by intervals and these intervals can impart some important information about the solution. Consider the following example.

Example 1 Show that is a solution to for Solution

Initial Condition(s) Initial Condition(s) are a condition, or set of conditions, on the solution that will allow us to determine which solution that we are after. Initial conditions (often abbreviated i.c.’s when I’m feeling lazy…) are of the form, So, in other words, initial conditions are values of the solution and/or its derivative(s) at specific points. As we will see eventually, solutions to “nice enough” differential equations are unique and hence only one solution will meet the given conditions. The number of initial conditions that are required for a given differential equation will depend upon the order of the differential equation as we will see.

Example 2 is a solution to , Solution and

Initial Value Problem An Initial Value Problem (or IVP) is a differential equation along with an appropriate number of initial conditions. Example 3 and , is an IVP Example 4 , is another IVP

2 First-order Ordinary Differential Equation A differential equation is an equation that contains the derivatives or differentials of one or more dependent variables with respect to one or more independent variables. If the equation contains only ordinary derivatives with respect to a single independent variable , the equation is called an ordinary differential equation , abbreviated by ODE We can write any ordinary differential equation in the following form. M(x,y)dx + N(x,y)dy = 0 M(x,y) and N(x,y) are functions of x and y.

3 Separation of Variables ODE form. M(x,y)dx + N(x,y)dy = 0 we will look at is separable differential equations. A separable differential equation is any differential equation that we can write in the following form. A(x)dx + B(y)dy = 0 A(x) is a function of only x and B(y) is a function of only y. Solving separable differential equation is fairly easy. We will integrate both sides. A(x)dx + B(y)dy = c

Example 5 Solve the following differential equation y dx + x2 dy = 0 Solution

Example 6 Solve the initial-value problem. 3y dx - 2x dy = 0 , y(2) = 1 Solution

Example 7 Solve the initial-value problem. (4x2 + 3)2 y / - 4xy2 = 0 , y(0) = 3/2 Solution

Example 8 Solve differential equation. , e is a constant dt = e (t sin x dx - cos x dt) Solution

4 Homogeneous Equations Definition A first-order differential equations that can be written in the form M(x,y)dx + N(x,y)dy = 0 where M(tx,ty) = tn M(x,y) and N(tx,ty) = tn N(x,y) is called a Homogeneous Differential Equation (of degree n).

Example 9 Show that the equation (x2 + yx)dx – y2dy = 0 is homogeneous. Solution

Solving Homogeneous Equations Homogeneous equation are reduced to separable equations by either of the substitutions y = ux or x = vy Use the substitution y = ux if N(x,y) is less complicated than M(x,y) and use x = vy if M(x,y) is less complicated than N(x,y).

Example 10 Solve the equation (x2 + y2)dx – 2xydy = 0 Solution

Example 11 Solve the initial-value problem. (y2 + 7xy+16x2)dx + x2dy = 0 , y(1) = 1 Solution

Example 12 Solve the initial-value problem. (x cos2(y/x) - y)dx + xdy = 0 , y(1) =  /4 Solution

Example 13 Solve the equation. y dx = (x +  y2 – x2)dy Solution

Do Activities ACTIVITY EXERCISES 1 (Separation of Variables) 1. Solve the initial-value problem (1  x 3 )dy  x 2 ydx  0 , y(1)  2 2. Solve the initial-value problem 2y cos xdx  sin xdy  0 , y(  )  2 2 ACTIVITY EXERCISES 2 (Homogeneous Equations) 1. Determine if the differential equation is homogeneous If so, determine its degree and solve its. [ x cos 2 ( y )  y]dx  xdy  0 x 2. Solve the initial-value problem y(3x  2y)dx  x2dy  0 , y(1)  2 Upload files and submit them to Google Classroom from 22-28 October 2020.

Review Separable equation M(x,y)dx + N(x,y)dy = 0 is ODE. We will look at is separable differential equations in the form, A(x)dx + B(y)dy = 0. After that take integral sign both sides. Homogeneous equation First, check degree of the equation in x and y, if the sum of exponents of the variables in each term is of the same degree. ( M(x,y) and N(x,y) are both homogeneous functions of the same degree. After that we create a new variable y = ux or x = vy and solve equation. Finally, we can then return to x and y by substituting u = y or v = x xy

5 Exact Equations Definition A first-order differential equations M(x,y)dx + N(x,y)dy = 0 is exact if and only if My  Nx

Example 14 Show that the equation 2xy3 dx + (1+3x2y2)dy = 0 is exact and that the equation x 2y dx + 5xy2 dy = 0 is not exact.

Solving the Exact Differential Equation M(x,y)dx + N(x,y)dy = 0 1. Assume that M(x,y)  xf and N(x,y)  yf 2. Integrate M(x,y) with respect to x. (Add an arbitrary function of y,g(y).) A similar algorithm can be stated so that in step 2, N(x,y) is integrated with respect to y as we show in Example 16 3. Differentiate the result in step 2 with respect to y and set the result equal to N(x,y). Solve for y g(y) 4. Integrate y g(y) with respect to y to obtain an expression for g(y). (There is no need to include an arbitrary constant.). 5. Substitute g(y) into the result obtained in step 2 for f(x,y). 6. A general solution is f(x,y) = C , where C is a constant. 7. Apply the initial condition if given..

Example 15 Solve y(2xy2 -3) + (3x2y2 – 3x+4y)y / = 0

Example 16 Solve (ysec2x + sec x tan x)dx + (tan x + 2y)dy = 0

Example 17 Solve the initial-value problem dy  xy2 1 , y(0) = 1 dx 1x2y

6 Linear Equations Definition A general solution of the linear first-order differential equations dy  p(x) y  q(x) is found by solving dx ddx [(x) y]  (x)q(x) for y where (x)  e p(x)dx is called an integrating factor for the linear equation.

Example 18 Find a general solution of x dy  y  x sinx. dx



Example 19 Solve the initial-value problem dy  5x4y  x4 , y(0) = -7 dx



Do Activities ACTIVITY EXERCISES 3 (Exact Equation) 1. Solve Exact Equation 2x(3x  y  yex2 )dx  (x2  3y2  ex2 )dy  0 ACTIVITY EXERCISES 4 (Linear Equation) 1. Solve Linear Equation 2(y  4x2 )dx  xdy  0 Upload files and submit them to Google Classroom from 29 Oct. – 3 Nov. 2020.