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Find the general solution of the given higher-order differential equation: y^{4} + y^{3} + y^{2} = 0

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.

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