Solve the equations by variation of parameters. y^''-y=e^x |...

Name: Problem 23, Chapter 17: Thomas' Calculus: Early Transcendentals in SI Units
Uploaded: 2022-01-24T16:33:22-08:00
Duration: 9 min 3 s
Channel: Nadir Musofer
Description: Solve the equations by variation of parameters. y^''-y=e^x

Key Concepts

Variation of Parameters

This is a method used to find a particular solution to a non-homogeneous linear differential equation. It involves using the solutions of the homogeneous equation to construct a formula where the constants are replaced by functions, and these functions are determined by solving a system derived from the original differential equation.

Fundamental Set of Solutions

This concept refers to a pair of linearly independent solutions to the homogeneous differential equation. These form the basis to express the general solution of the homogeneous equation, allowing any solution to be written as a linear combination of these independent functions.

Wronskian Determinant

The Wronskian is a determinant used to establish the linear independence of functions. In the context of second-order differential equations, a non-zero Wronskian on an interval indicates that the chosen fundamental set of solutions is linearly independent, which is crucial for methods such as variation of parameters.

Second-Order Linear Differential Equations

These are differential equations involving the second derivative of an unknown function and can typically be written in the form a(x)y'' + b(x)y' + c(x)y = f(x). The study of these equations forms a critical part of differential equations, as they model various phenomena in physics and engineering.

Homogeneous Solution

For a linear differential equation, the homogeneous solution is the solution to the corresponding equation when f(x) equals zero. Finding the homogeneous solution typically involves solving the characteristic equation, which provides the basis for the general solution by offering a fundamental set of solutions.

Transcript

00:01 Dear students here we have a differential equation and we have to use variation of parameter to solve this differential equation so dear students so dear students this is we know that this is the non -homogeneous differential equation so this will have a particular integral and a complementary function in its general solution so we have to find the complementary function and the particular integral so dear students here we have we have the i have mentioned the working process here so dear student this is the this is the standard form of the equation so here in our perspective what is what is r it is e to the power x so here from this we will find the auxiliary equation and from auxiliary equation we will find the complementary function and dear students we will use roscanian of you and v for a particular integral so dear students let's find first we have to find the auxiliary equations so for before the auxiliary equation this will be the differential equation can be written as as d square y minus y is equal to e to the power x so here if we take y as a common so this will give d student this will give d square minus 1 into y is equal to e to the power x so here d students to find the auxiliary equation so this will be m square minus 1 so d students here notice that it is this will be so d student this will be this will be this will be m square minus 1 is equal to 0 and from here this will be equal to m is equal to 1 m square is equal to 1 so m will be equal to plus minus 1 so d students here we have a different different roots so the complementary function will be equal to y c will be equal to d students it c1 this is c1 e to the power x plus c2 c2 e to the power minus x so d students here now we have to find the particular integral so for for particular integral we have u of f of x and plus v of x so d s students here here in this our perspective we we have u is equal to u is this one here we have i have already told you here this is u with c1 so with c1 this is e to the power x and v is what here is dea students it's a to power minus x so d s students here we have to find f of x and g of x so for f of x and g of x we need the roscanian of u of x we need the roscanian of u and v so the roscanian is equal to u v and u you and the derivative of u and the derivative of v so by putting the values we will get e to the power x a to the power minus x d s students and what is this one the prime and the derivative of e to the power x is e to the power x and minus e to the power minus x so d s students here so d students the term using the determinant formula this will give this will give d students this will give minus 1 and here it will be also minus 1 because we know that if you multiply these both this and this so by using the concept of the using the concept of same with same base the powers will be aid added so de students we will use and if we add them we will get e to the power 0 and e to the power 0 something power 0 is 1 that's why i have here mentioned i've written here minus 1 mean minus 1…

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