RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE HADAPSAR, PUNE

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RAYAT SHIKSHAN SANSTHA’S S. M. JOSHI COLLEGE HADAPSAR, PUNE
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RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE HADAPSAR, PUNE-411028. PRESANTATION BY Prof . DESAI S.S Mathematics department Subject – differential equations

Differential Equations An equation which involves unknown function and its derivatives 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

Differential Equations A linear differential equation of order n is a differential equation written in the following form: where an(x) is not the zero function General solution : looking for the unknown function of a differential equation Particular solution (Initial Value Problem) : looking for the unknown function of a differential equation where the values of the unknown function and its derivatives at some point are known Issues in finding solution : existence and uniqueness

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: Solution : 1. Multiply the equation by integrating factor: 2. The variable are separated : 3. Integrating to find the solution:

1st Order DE - Homogeneous Equations Homogeneous Function f (x,y) is called homogenous of degree n if : Examples:  homogeneous of degree 4  non-homogeneous

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 : 2. The equation become separable equation: 3. Use solution method for separable equation 4. After integrating, v is replaced by y/x

1st Order DE – Exact Equation The differential equation M(x,y)dx + N(x,y)dy = 0 is an exact equation if : The solutions are given by the implicit equation where : F/ x = M(x,y) and F/ y = N(x,y) Solution : 1. Integrate either M(x,y) with respect to x or N(x,y) to y. Assume integrating M(x,y), then : 2. Now : or :

1st Order DE – Non Exact Equation The differential equation M(x,y)dx + N(x,y)dy = 0 is a non exact equation if : The solutions are given by using integrating factor to change the equation into exact equation Solution : 1. Check if : then integrating factor is or if : then integrating factor is

1st Order DE – Non Exact Equation 2. Multiply the differential equation with integrating factor which result an exact differential equation 3. Solve the equation using procedure for an exact equation

1st Order DE – Linear Equation A first order linear differential equation has the following general form: Solution : 1. Find the integrating factor: 2. Evaluate : 3. Find the solution:

2nd Order DE – Linear Equation A second order differential equation is an equation involving the unknown function y, its derivatives y' and y'', and the variable x. We will consider explicit differential equations of the form : A linear second order differential equations is written as: When d(x) = 0, the equation is called homogeneous, otherwise it is called nonhomogeneous

2nd Order DE – Linear Equation To a nonhomogeneous equation (NH) we associate the so called associated homogeneous equation (H) Main result: The general solution to the equation (NH) is given by: where: (i) yh is the general solution to the associated homogeneous equation (H); (ii) yp is a particular solution to the equation (NH).

2nd Order DE – Linear Equation Basic property Consider the homogeneous second order linear equation or the explicit one Property: If y1 and y2 are two solutions, then: is also a solution for any arbitrary constants c1 ,c2

2nd Order DE – Reduction of Order Reduction of Order Technique This technique is very important since it helps one to find a second solution independent from a known one. Let y1 be a non-zero solution of: Then, a second solution y2 independent of y1 can be found as: Where: The general solution is then given by

2nd Order DE – Homogeneous LE with Constant Coefficients Homogeneous Linear Equations with Constant Coefficients A second order homogeneous equation with constant coefficients is written as: where a, b and c are constant The steps to follow in order to find the general solution is as follows: (1) Write down the characteristic equation This is a quadratic. Let 1 and 2 be its roots we have

2nd Order DE – Homogeneous LE with Constant Coefficients (2) If 1 and 2 are distinct real numbers (if b2 - 4ac > 0), then the general solution is: (3) If 1 = 2 (if b2 - 4ac = 0), then the general solution is: (4) If 1 and 2 are complex numbers (if b2 - 4ac < 0), then the general solution is: Where:

2nd Order DE – Non Homogeneous LE To a nonhomogeneous equation (NH) we associate the so called associated homogeneous equation (H) Main result: The general solution to the equation (NH) is given by: where: (i) yh is the general solution to the associated homogeneous equation (H); (ii) yp is a particular solution to the equation (NH).

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