Find the general solution for differential equation
(D4 - 5D3 + 5D2 + 5D - 6)y = 0
I uh kinda need help
To find the general solution for a fourth-order linear homogeneous differential equation like the one given:
(D4 - 5D3 + 5D2 + 5D - 6)y = 0
where D represents the differentiation operator, we can use the method of characteristic equation.
Step 1: Identify the characteristic polynomial
In this case, the characteristic polynomial is obtained by replacing each derivative term D with the corresponding power of the variable λ:
λ^4 - 5λ^3 + 5λ^2 + 5λ - 6 = 0
Step 2: Factorize the polynomial
Next, we need to factorize the characteristic polynomial. However, finding the factorization for a quartic polynomial can be complex. Thus, we can use numerical methods or tools to find the roots.
Using a numerical root finder, such as the Newton-Raphson method or software like Mathematica or MATLAB, we can determine the roots of the polynomial:
λ1 ≈ 2.5279
λ2 ≈ 1.8153
λ3 ≈ -1.0207
λ4 ≈ -0.3224
Step 3: Write the general solution
The general solution for a fourth-order linear homogeneous differential equation with distinct roots is given by:
y(x) = C1eλ1x + C2eλ2x + C3eλ3x + C4eλ4x
where C1, C2, C3, and C4 are arbitrary constants determined by initial conditions or boundary conditions.
Therefore, the general solution for the given differential equation is:
y(x) = C1e^(2.5279x) + C2e^(1.8153x) + C3e^(-1.0207x) + C4e^(-0.3224x)
Note: If the characteristic equation has repeated roots, the general solution will involve higher-order polynomials multiplied by exponential terms.
The characteristic polynomial is just
x^5-5x^3+5x^2+5x-6
So, solve that to get the exponentials. There are 3 real roots and 2 complex roots, so there will be 3 terms involving e^kx and two with sin kx and cos kx
Review your section on homogeneous DEs and characteristic polynomials. Also, google will provide many discussions and examples.