Determine the general solution to the given differential equation. Derive your trial solution us

Determine the general solution to the given differential...

Key Concepts

Linear Differential Equations with Constant Coefficients

Linear differential equations involve the dependent variable and its derivatives appearing in a linear form. When the coefficients are constant, the methods of solution become systematic and mainly rely on algebraic techniques, such as finding the roots of the characteristic equation.

Differential Operator Notation

Differential operator notation uses symbols, like D, to denote differentiation, thereby simplifying the representation of differential equations. This abstraction facilitates the use of algebraic methods, including factorization and the application of the annihilator method.

Characteristic Equation and Complementary Solution

The complementary solution is the general solution to the homogeneous part of the differential equation, obtained by setting the non-homogeneous term to zero. It is found by translating the differential equation into an algebraic 'characteristic equation', solving for its roots, and then writing the solution based on these roots.

Particular Solution

The particular solution is a specific solution that accounts for the non-homogeneous part of the differential equation. It is determined by methods such as undetermined coefficients or the annihilator technique, and when combined with the complementary solution, forms the general solution of the equation.

Annihilator Technique

The annihilator technique involves finding a differential operator that, when applied to the non-homogeneous term, yields zero. By applying this operator to the entire equation, the problem is transformed into a homogeneous differential equation, which can be solved to guide the selection of a suitable trial solution for the particular part.

Transcript

00:01 Okay, so in this problem, let's start by finding the solution to the corresponding homogeneous equation.

00:09 So that's going to be d squared minus 1 times the complementary solution is equal to 0.

00:15 So then our corresponding polynomial p of r is going to be r squared minus 1.

00:21 You don't need to put that in parentheses.

00:23 Square minus 1 is equal to 0.

00:25 So then we just get that r is going to be equal to plus or minus 1.

00:30 Here.

00:32 So our complementary solution, y of c of x, is going to be equal to c1e to the x plus c2e to the negative x.

00:48 So now for our right -hand side, we have f of x is equal to 3e to 2x minus 8e to the 3x.

00:58 Here we have an r is equal to 2, and then here we have an r is equal to 3.

01:02 So our annihilator, a1d, is going to be equal to first d minus r1, which is d minus 2, and then annihilator 2 is going to be equal to d minus r2, which is 3.

01:15 So our total annihilator, a of d, is going to be equal to d minus 2 times d minus 3.

01:23 If we apply this annihilator to both sides or both sides of this equation, we have d minus 2, d minus 3 times d squared minus 1.

01:34 Of y is equal to and then this makes this right -hand side equal to zero.

01:41 So this has the corresponding polynomial, r minus two, r minus three, and then r squared minus one.

01:52 And we already know the roots of this one are going to be plus or minus one.

01:56 And here we get roots of two and three.

01:58 So our general solution is going to be now our complementary solution, c1e to the x, plus c2e to the negative x, and then a plus, then we'll have an a0, e to the negative 2x, plus an a1, e to the 2x, plus an a1e to the 3x.

02:27 This second half here is going to be our trial solution, yp of x.

02:35 Now to solve for a not an a1, we're going to plug this into our solution here, or our equation here.

02:43 So we're going to have d squared minus 1 of a knot e to the 2x plus a 1e to the 3x.

02:53 And we're going to set this equal to our right hand side 3e to the 2x minus 8e to the 3x.

03:00 So first we apply this to this.

03:05 So we're going to take d and then d of 3.

03:10 D of this, sorry, of this.

03:14 So first d of this is going to be 2a0 e to the 2x plus 3a1 e to the 3x.

03:25 And then we also subtract off that because of this minus 1 here...