First order linear differential equation
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1. A differential equation relates an unknown function and its derivatives, and can be ordinary (involving one variable) or partial (involving partial derivatives). 2. Linear differential equations have dependent variables and derivatives that are of degree one, and coefficients that do not depend on the dependent variable. 3. Common methods for solving first-order linear differential equations include separation of variables, homogeneous equations, and exact equations.
First order linear differential equation
- 3. Differential Equations An equation which involves unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. ordinary differential equation (ode) : not involve partial derivatives partial differential equation (pde) : involves partial derivatives order of the differential equation is the order of the highest derivatives Examples: second order ordinary differential equation first order partial differential equation 2 2 3 sin d y dy x y dxdx y y x t x t x x t
- 4. Terminologies In Differential Equation • Existence: Does a differential equation have a solution? • Uniqueness: Does a differential equation have more than one solution? If yes, how can we find a solution which satisfies particular conditions? • A problem in which we are looking for the unknown function of a differential equation where the values of the unknown function and its derivatives at some point are known is called an initial value problem (in short IVP). • If no initial conditions are given, we call the description of all solutions to the differential equation the general solution.
- 5. Differential Equations Some Application of Differential Equation in Engineering
- 6. Linear Differential Equation A differential equation is linear, if 1. dependent variable and its derivatives are of degree one, 2. coefficients of a term does not depend upon dependent variable. Example: 36 4 3 3 y dx dy dx yd is non - linear because in 2nd term is not of degree one. .0932 2 y dx dy dx ydExample: is linear. 1. 2.
- 7. ( , )y f x y
- 8. First Order Linear Equations • A linear first order equation is an equation that can be expressed in the form Where P and Q are functions of x
- 9. History YEAR PROBLEM DESCRIPTION MATHAMATICIAN 1690 Problem of the Isochrones Finding a curve along which a body will fall with uniform vertical velocity James Bernoulli 1728 Problem of Reducing 2nd Order Equations to 1st Order Finding an integrating factor Leonhard Euler 1743 Problem of determining integrating factor for the general linear equation Concept of the ad- Joint of a differential equation Joseph Lagrange 1762 Problem of Linear Equation with Constant Coefficients Conditions under which the order of a linear differential equation could be lowered Jean d’Alembert
- 10. Methods Solving LDE 1. Separable variable M(x)dx + N(y)dy = 0 2. Homogenous M(x,y)dx+N(x,y)dy=0, where M & N are nth degree 3. Exact M(x,y)dx + N(x,y)dy=0, where M/ðy=0, where ðM/ðy = ðN/ðx
- 11. Solution of Differential Equation
- 12. 1st Order DE - Separable Equations The differential equation M(x,y)dx + N(x,y)dy = 0 is separable if the equation can be written in the form: 02211 dyygxfdxygxf Solution : 1. Multiply the equation by integrating factor: ygxf 12 1 2. The variable are separated : 0 1 2 2 1 dy yg yg dx xf xf 3. Integrating to find the solution: Cdy yg yg dx xf xf 1 2 2 1
- 13. 1st Order DE - Homogeneous Equations Homogeneous Function f (x,y) is called homogenous of degree n if : y,xfy,xf n Examples: yxxy,xf 34 homogeneous of degree 4 yxfyxx yxxyxf , , 4344 34 yxxyxf cossin, 2 non-homogeneous yxf yxx yxxyxf n , cossin cossin, 22 2
- 14. 1st Order DE - Homogeneous Equations The differential equation M(x,y)dx + N(x,y)dy = 0 is homogeneous if M(x,y) and N(x,y) are homogeneous and of the same degree Solution : 1. Use the transformation to : dvxdxvdyvxy 2. The equation become separable equation: 0,, dvvxQdxvxP 3. Use solution method for separable equation Cdv vg vg dx xf xf 1 2 2 1 4. After integrating, v is replaced by y/x
- 15. 1st Order DE – Exact Equation The differential equation M(x,y)dx + N(x,y)dy = 0 is an exact equation if : Solution : The solutions are given by the implicit equation x N y M CyxF , 1. Integrate either M(x,y) with respect to x or N(x,y) to y. Assume integrating M(x,y), then : where : F/ x = M(x,y) and F/ y = N(x,y) ydxyxMyxF ,, 2. Now : yxNydxyxM yy F ,', or : dxyxM y yxNy ,,'
- 16. 1st Order DE – Exact Equation 3. Integrate ’(y) to get (y) and write down the result F(x,y) = C Examples: 1. Solve : 01332 3 dyyxdxyx Answer:
- 17. Newton's Law of Cooling • It is a model that describes, mathematically, the change in temperature of an object in a given environment. The law states that the rate of change (in time) of the temperature is proportional to the difference between the temperature T of the object and the temperature Te of the environment surrounding the object. d T / d t = - k (T - Te) Let x = T - Te so that dx / dt = dT / dt d x / d t = - k x The solution to the above differential equation is given by x = A e - k t substitute x by T – Te T - Te = A e - k t Assume that at t = 0 the temperature T = To
- 18. T0 - Te = A e o which gives A = To-Te The final expression for T(t) is given by T(t) = Te + (To- Te) e - k t This last expression shows how the temperature T of the object changes with time.
- 19. Growth And Decay • The initial value problem where N(t) denotes population at time t and k is a constant of proportionality, serves as a model for population growth and decay of insects, animals and human population at certain places and duration. Integrating both sides we get ln N(t)=kt+ln C or or N(t)=Cekt C can be determined if N(t) is given at certain time. )( )( tkN dt tdN kdt tN tdN )( )(
- 20. Carbon dating Let M(t) be the amount of a product that decreases withtime t and the rate of decrease is proportional to the amount M as follows d M / d t = - k M where d M / d t is the first derivative of M, k > 0 and t is the time. Solve the above first order differential equation to obtain M(t) = Ae-kt where A is non zero constant. It we assume that M = Mo at t = 0, then M= Ae0 which gives A = Mo The solution may be written as follows M(t) = Mo e-kt
- 21. Economics and Finance • The problems regarding supply, demand and compounding interest can be calculated by this equation is a separable differential equation of first-order. We can write it as dP=k(D-S) dt. Integrating both sides, we get P(t)=k(D-S)t+A where A is a constant of integration. Similarly S(t)=S(0) ert ,Where S(0) is the initial money in the account )( SDk dt dP