1-10 Solve the differential equation or initial-value problem using the method of undetermined c

1-10 Solve the differential equation or initial-value...

Key Concepts

Linear Differential Equations

Linear differential equations involve functions and their derivatives, and have the property that the dependent variable and its derivatives appear to the first power and are not multiplied together. In equations with constant coefficients, the formulas for the solution can be systematically determined using algebraic methods, which makes them more tractable for analysis and solution techniques, including the method of undetermined coefficients.

Characteristic Equation

The characteristic equation is derived from the homogeneous part of a linear differential equation with constant coefficients by assuming a solution of the exponential form. The roots of this polynomial equation determine the structure of the homogeneous (complementary) solution. Real and complex roots lead to different forms of the solution, such as exponentials, sines, and cosines.

Method of Undetermined Coefficients

This method is used to find a particular solution to non-homogeneous linear differential equations. It involves making an educated guess for the form of the particular solution based on the type of function present on the right-hand side (such as polynomials or exponentials) and then determining the unknown coefficients by substituting the guess into the differential equation.

Complementary and Particular Solutions

The overall solution to a non-homogeneous differential equation is obtained by combining the complementary (homogeneous) solution, which solves the equation without the non-homogeneous part, with the particular solution found via a method like undetermined coefficients. This approach uses the superposition principle, as the sum of any solution to the homogeneous equation and a particular solution to the non-homogeneous equation is also a solution to the full equation.

Transcript

00:01 So let's start this problem out by solving for the homogenous solution.

00:03 So i have r squared minus 2r plus 2 equals to 0.

00:10 And i do not see any immediate factoring come to mind.

00:13 So i'm just going to use a quadratic formula.

00:15 So i'm going to have negative b plus or minus a square of b squared minus 4 times a times c all over 2 times a.

00:32 And that equals to 1 plus or minus.

00:37 The square root of 4, sorry, negative 4 divided by 2, and that equals to 1 plus or minus i.

00:52 Right, and so r equals to 1 plus or minus i, and with that we can build a homogenous solution.

01:00 So we'll have c1, e to the x, cosine x, plus c2, e to the x, sine x.

01:11 And our guess for the particular solution is going to be a x plus b plus c e to the x and we're going to have to take the derivative twice so the first derivative is going to be a plus c e to the x and our second derivative is going to be c e to the x so let's plug that into our original equation and let's remind ourselves that the original what's y double prime minus 2y prime plus 2y equals to x plus e to the x? so when we plug that in, y double prime is c e to the x minus 2 times a minus 2c e to the x plus 2ax plus 2b plus 2c e to the x that all equals x plus e to the x that all equals x plus e to the x.

02:26 And so let's try and create a system of equations here.

02:30 So we'll have, let's actually simplify this down, actually.

02:35 Let's see.

02:38 So we have these terms right here that are alike...

Please give Ace some feedback