This problem aims to find the differential of a higher-order polynomial whose equation is given. An expert understanding of higher-order equations and quadratic formulas is required to solve this problem which is explained below:
This is called a homogeneous linear differential equation with constant coefficients, so we’ll start by writing down the characteristic equation that’s of the order four: y^ {4} + y^ 3+ y^ 2 = 0
We can use complex exponential functions or use trigonometric functions for complex distinct roots.
The general solution using trigonometric function is:
\[ y = c_1 cos(2t) + c_2 sin(2t) + c_3t cos(2t) + c_4t sin(2t) \]
where $c_1, c_2, c_3, c_4$ are free variables.
The general solution using complex exponential function is:
\[ y = C_1 e^ {2it} + C_2t e^ {2it} + C_3 e^ {-2it} + C_4t e^ {-2it} \]
where $C_1, C_2, C_3, C_4$ are free variables.
Expert Answer
The first step is to find the roots of this equation. To solve this, we will factor out $y^ 2$, taking $y^ 2$ common:
\[ y^ 2 ( y^ {2} + y+ 1) = 0 \]
Putting $y^2$ equals to $0$ leaves us with $2$ equations:
$y = 0$ with multiplicity of $2$ and $ ( y^ {2} + y+ 1) = 0$.
Solving the remaining $ ( y^ {2} + y+ 1) $ equals to $0$ using the quadratic formula:
\[ y^ {2} + y+ 1 = 0 \]
First, the quadratic formula is given as:
\[ y = \dfrac{-b \pm \sqrt {b^ 2 – 4ac}} {2a} \]
Putting $a = 1, b = 1$ and $c = 1$ in the formula gives us:
\[ y = \dfrac{-1 \pm \sqrt {1 – 4} }{2} \]
\[ y = \dfrac{-1}{2} \pm \dfrac{i \sqrt {3} }{2} \]
Thus, the final roots are $0, 0, \left( \dfrac{-1}{2} + \dfrac{i \sqrt {3} }{2} \right) and \left( \dfrac{-1}{2} – \dfrac{i \sqrt {3} }{2} \right)$
We will use the complex exponential formula for our general solution:
\[ y = C_1 e^ {2it} + C_2t e^ {2it} + C_3 e^ {-2it} + C_4t e^ {-2it} \]
The general solution becomes:
\[ y = C_1 e^ {0x} + C_2 xe^ {0x} + C_3 e^ {\dfrac{-x}{2}} cos \left( \dfrac {\sqrt{3}}{2}x \right) + C_4 e^ {\dfrac{-x}{2}} sin \left( \dfrac {\sqrt{3}}{2}x \right) \]
Numerical Result
\[ y = C_1 + C_2 x + C_3 e^{\dfrac{-x}{2}} cos \left( \dfrac {\sqrt{3}}{2}x \right) + C_4 e^{\dfrac{-x}{2}} sin \left( \dfrac {\sqrt{3}}{2}x \right) \]
Example
For the given higher-order differential equation, solve for the general solution:
\[ y^{4} + 8y” + 16y = 0 \]
Solving for $y$, we get:
\[ y^{4} + 8y^2 + 16y = 0 \]
\[ (y^ 2 + 4)^2 = 0 \]
The roots are $2i, 2i, -2i, -2i$. Thus, we have repeated roots.
So the general solution becomes:
\[ y= C_1 e^ {2ix} + C_2 xe^{2ix} + C_3x e^ {-2ix} + C_4 e^ {-2ix} \]
One thing to note here is that the method of characteristic roots does not work for linear polynomial equations with variable coefficients.