Solving the Differential Equation D3r36D2r211D?y0 and D3r36D2r212D?y0 Through Characteristic Equations

In this article, we will delve into the process of solving two differential equations using characteristic equations. We will begin by discussing the general method and then proceed to solve the given differential equations step-by-step. This approach is crucial for understanding and solving higher-order differential equations in various fields of mathematics and physics.

Solving D3r36D2r211D?y0

To find the general solution of the differential equation D3r36D2r211D?y0, we first need to determine the characteristic polynomial associated with the differential operator D, where D represents differentiation with respect to x.

The characteristic polynomial is given by:

1. Determining the Characteristic Polynomial

The characteristic polynomial is:

r36r211r60

We can factor this polynomial or use the Rational Root Theorem to find its roots. Testing possible rational roots, we find that r-1 is a root. We can then perform polynomial long division to factor r36r211r6 by r 1.

Step 1: Polynomial Long Division

Divide the leading term: r3/r r2. Multiply r2 by r 1: r3 r2. Subtract: r36r211r6-(r3 r2)5r211r6. Divide the leading term: 5r2/r 5r. Multiply 5r by r 1: 5r25r. Subtract: 5r211r6-(5r25r)6r6. Divide the leading term: 6r/r 6. Multiply 6 by r 1: 6r6. Subtract: 6r6-6r60.

Thus, we have:

r36r211r6 (r 1)(r25r6)

Step 2: Factor the Quadratic

We factor the quadratic r25r6 as follows:

r25r6 (r2)(r3)

Step 3: Roots of the Characteristic Equation

The complete factorization of the characteristic polynomial is:

(r 1)(r2)(r3)0

This gives us the roots:

r?-1 r?-2 r?-3

Step 4: General Solution

The general solution y(x) of the differential equation D3r36D2r211D?y0 is given by:

y(x) C?e-x C?e-2x C?e-3x

where C?, C? and C? are constants determined by initial conditions.

Solving D3r36D2r212D?y0

For the differential equation D3r36D2r212D?y0, the characteristic equation is given by:

D36D212D80

We can simplify the characteristic equation as follows:

D36D212D8 D32?D20.5?D1? (D2)((D0.5)31?) (D2)(D0.53) (D2)(D0.51)(D0.51) (D2)(D1.5)(1.5)

From the simplified form, we see that the roots are:

D -2 D -2 D -2

Step 2: General Solution

The general solution y(t) of the differential equation is given by:

y(t) c?t c?t2 c?e-2t

Conclusion

By following the steps we outlined, we were able to find the general solutions for both differential equations. Understanding these methods is essential for solving higher-order differential equations and applying them in various fields of mathematics and physics. The use of characteristic equations and polynomial long division provides a systematic approach to finding the solutions.

Keywords

Differential equation, characteristic polynomial, solution, characteristic equation