Question

Find the general solution. $$ 27 y^{\prime \prime \prime}+27 y^{\prime \prime}+9 y^{\prime}+y=0 $$

Short answer

Question: Given the third-order linear homogeneous differential equation 27y''' + 27y'' + 9y' + y = 0, find the general solution. Answer: The general solution for the given third-order linear homogeneous differential equation is y(t) = e^(-1/3t) * [C1 * cos(1/3t) + C2 * sin(1/3t)] + C3 * e^(-1/3t), where C1, C2, and C3 are constants determined by the initial or boundary conditions.

Step-by-step solution

Obtain the Characteristic Equation

First, we will create the characteristic equation for the given differential equation. The characteristic equation will have the form: $$ ar^3 + br^2 + cr + d = 0 $$ where a, b, c, and d are coefficients of the differential equation, and r is the roots of the characteristic equation. Now plug in the coefficients from the given differential equation: $$ 27r^3 + 27r^2 + 9r + 1 = 0 $$

Solve the Characteristic Equation

To solve the characteristic equation, we need to find the roots, which will be used to construct the general solution of the given differential equation. Factoring out 27 from all the coefficients, we get: $$ (27r^3 + 27r^2 +9r +1) = 27(r^3 + r^2+\frac{1}{3}r + \frac{1}{27}) = (3r+1) \cdot (9r^2 + 1) = 0 $$ We can now solve for r: $$ 3r + 1 = 0 \rightarrow r = -\frac{1}{3} \qquad \text{(the solution of first-factor)} $$ $$ 9r^2 + 1 = 0 \rightarrow r^2 = -\frac{1}{9} $$ The solution of the second-factor are complex conjugate roots: $$ r_{1,2} = \pm \frac{i}{3} $$

Construct the General Solution

We now have the three roots (-1/3, i/3, and -i/3), and we will use them to construct the general solution for the given differential equation. Since we have two complex conjugate roots and one real root, the general solution will have the form: $$ y(t) = e^{r_1t} \cdot [C_1 \cos(\omega t) + C_2 \sin(\omega t)] + C_3e^{r_3t} $$ Here, \(r_1=\frac{i}{3}\), \(\omega = |r_1|=\frac{1}{3}\), \(r_3=-\frac{1}{3}\), and \(C_1, C_2, C_3\) are constants to be determined by initial/boundary conditions. Plugging in the values of roots, we get the general solution: $$ y(t) = e^{\frac{-1}{3}t} \cdot [C_1 \cos(\frac{1}{3}t) + C_2 \sin(\frac{1}{3}t)] + C_3e^{\frac{-1}{3}t} $$ This is the general solution for the given third-order linear homogeneous differential equation.

Key concepts

Characteristic Equation

When dealing with differential equations, especially linear homogeneous ones, a characteristic equation is a powerful tool. It's a polynomial equation derived from the differential equation itself. If you have a differential equation like:\[ a y^{(n)} + b y^{(n-1)} + ext{...} + z y = 0 \]The characteristic equation translates this into a polynomial in terms of \( r \):\[ ar^n + br^{n-1} + ext{...} + z = 0 \]You obtain this by substituting \( y = e^{rt} \) into the differential equation. Solving the characteristic equation gives us the roots \( r \), which are crucial for finding the general solution. For our specific problem:\[ 27r^3 + 27r^2 + 9r + 1 = 0 \]we solve this cubic polynomial to find the roots which indicate the nature of our solution in terms of real and complex exponentials.

Complex Roots

Complex roots play a significant role in determining the form of the solution to differential equations. When the characteristic equation yields complex roots like \( r = rac{i}{3} \) and \( r = -rac{i}{3} \), this results in sinusoidal solutions involving sine and cosine functions. Complex roots typically appear in conjugate pairs. For a conjugate pair like \( a \pm bi \), the solution includes terms of the form:\[ e^{at}(C_1 \cos(bt) + C_2 \sin(bt)) \]In our exercise, the solution includes :- \( \frac{i}{3} \) and \( -\frac{i}{3} \) indicate that the solution will involve \( e^{0t} \) (which simplifies to 1) and trigonometric functions \( \cos(\frac{1}{3}t) \) and \( \sin(\frac{1}{3}t) \). This occurs because the imaginary component forces the introduction of oscillatory behavior, represented by sine and cosine.

General Solution of Differential Equations

The general solution of a differential equation encompasses all possible solutions, tailored by the initial conditions of the problem. With the roots of the characteristic equation identified, one constructs the solution based on their nature:- Real roots contribute exponential terms- Complex roots contribute trigonometric termsFor our problem, the characteristic roots were \( -\frac{1}{3} \), \( \pm \frac{i}{3} \). Thus, the general solution:\[ y(t) = e^{\frac{-1}{3}t} (C_1 \cos(\frac{1}{3}t) + C_2 \sin(\frac{1}{3}t)) + C_3e^{\frac{-1}{3}t} \]This form reflects the mixed nature of the roots, demonstrating the blend of exponential decay and harmonic oscillations inherent to such solutions. The constants \( C_1, C_2, \) and \( C_3 \) will be defined by specific boundary conditions usually given in additional problem statements.