Solving the Fourth-Order Linear Homogeneous Differential Equation

The solution to the fourth-order linear homogeneous differential equation is a significant topic in advanced calculus and differential equations. This article will guide you through the process of solving the equation:

4D^4-20D^3 35D^2-25D 6y0,

where D represents the differential operator fracdod{x}. The solution involves finding the characteristic polynomial, factoring, and solving for the roots.

1. Formulating the Characteristic Polynomial

To begin, we rewrite the differential equation in terms of the characteristic polynomial. The differential operator D raised to the fourth power and linear combinations of it translate into the polynomial:

4r^4-20r^3 35r^2-25r 60.

2. Finding the Roots

The roots of this polynomial are essential for constructing the general solution. We use the Rational Root Theorem to test potential rational roots. The possible values to test are:

tpm1 tpm2 tpm3 tpm6 tpmfrac12; tpmfrac32; tpmfrac14; tpmfrac34;

Testing these values, we find:

Testing r -1: 4(-1)^4-20(-1)^3 35(-1)^2-25(-1) 64 20 35 25 60. Thus, r -1 is a root.

3. Polynomial Division

With r -1 as a root, we can perform polynomial division to simplify the equation. Dividing 4r^4-20r^3 35r^2-25r 6 by r 1, we get:

4r^4-20r^3 35r^2-25r 6r-1(4r^3-16r^2-19r-6).

4. Solving the Cubic Polynomial

Next, we solve the cubic polynomial 4r^3-16r^2-19r-60. Again, we apply the Rational Root Theorem. Testing r -frac12;: 4left(-frac12;right)^3-16left(-frac12;right)^2-19left(-frac12;right)-6 -frac12; - 4 frac{19}{2} - 6 0. Thus, r -frac12; is another root. Dividing 4r^3-16r^2-19r-6 by r frac12;, we get:

4r^3-16r^2-19r-6r frac12;left(4r^2-14r-12right).

5. Solving the Quadratic Polynomial

Finally, we solve the quadratic polynomial 4r^2-14r-120 using the quadratic formula:

rfrac{-bpmsqrt{b^2-4ac}}{2a}frac{-(-14)pmsqrt{(-14)^2-4cdot4cdot(-12)}}{2cdot4}frac{14pmsqrt{196 192}}{8}frac{14pm14}{8}.

This gives us:

rfrac{-12}{8}-frac{3}{2} quad text{and} quad rfrac{-16}{8}-2.

6. Summary of Roots

The roots of the characteristic equation are:

tr -1 with multiplicity 1 tr -frac12; with multiplicity 1 tr -frac{3}{2} with multiplicity 1 tr -2 with multiplicity 1

7. General Solution

The general solution of the differential equation is given by:

y(x) C_1 e^{-x} C_2 e^{-frac{1}{2}x} C_3 e^{-frac{3}{2}x} C_4 e^{-2x},

where C_1, C_2, C_3, and C_4 are arbitrary constants determined by initial or boundary conditions.