Determine a particular solution to the given differential...

Determine a particular solution to the given differential equation of the form $y_{p}(x)=A_{0} e^{-x} .$ Also find the general solution to the differential equation:
$$
y^{\prime \prime \prime}+5 y^{\prime \prime}+6 y^{\prime}=6 e^{-x}
$$

Key Concepts

Linear Differential Equations with Constant Coefficients

This concept involves differential equations where the coefficients of the derivatives are constant. Such equations are often solved through a combination of methods that leverage the linearity property, making it easier to separate the solution into a particular part and a homogeneous part.

Method of Undetermined Coefficients

This technique is used to find a particular solution to a nonhomogeneous linear differential equation by guessing a form for the solution based on the right-hand side of the equation. The coefficients in the guessed solution are then determined by substituting the guess into the differential equation and solving the resulting system.

Characteristic Equation

The characteristic equation is derived by replacing the derivative operators in the homogeneous version of the differential equation with a variable (often r). Solving this algebraic polynomial gives the roots, which are used to construct the general solution for the homogeneous part of the differential equation.

Superposition Principle

This principle states that the general solution of a linear differential equation is the sum of its homogeneous (complementary) solution and a particular solution. By combining these two parts, one obtains the complete general solution to the original nonhomogeneous differential equation.

Transcript

00:01 Okay, so we need to find a solution to this differential equation given this particular solution.

00:06 So we're going to need to find all the derivatives up to the third derivative of yp.

00:10 So we're going to get negative a, naught, e to the negative x.

00:14 Then yp double prime of x is just going to be a0, e to the negative x.

00:19 And then yp triple prime is going to be, again, negative a, knot, e to the negative x.

00:25 Substituting all of these in our equation here.

00:28 So we have negative a, naught, e to the negative.

00:30 X plus 5 a knot e to the negative x and then minus 6 a not e to the negative x is equal to 6 e to the negative x so we have 1 minus 5 or 1 plus 5 is 4 minus 6 is going to give us total negative 2 a not e to the negative x equal to 6 e to the negative x so then we need to set negative 2 equal to 6 so a not is going to be equal to negative 3.

01:04 So our particular solution, yp of x, is going to be equal to negative 3, e to the negative x.

01:14 So now to find the general solution, we first need to find the homogeneous solution.

01:18 So that's the solving of this differential equation here, y triple prime plus 5, i double prime plus 6 y prime is equal to 0.

01:26 We're going to assume solutions of the form y is equal to e to the rx, so y prime is equal to r sorry, r e to the rx, y double prime is equal to e to r squared, e to the rx, and then y triple prime is equal to r cubed, e to the rx...

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