The aim of this problem is to understand the** general solution** to the **higher order differential equations**.Â To solve such a question, we need to have a clear concept of **polynomial solution** and the **general solution** of the **differential equations**.

We basically convert the given **differential equation into an algebraic polynomial** by assuming that the **order of the differentiation is equivalent to the degree of the polynomial** of the normal algebraic expressions.

Having made the above assumption, we simply **solve the higher order polynomial** and the resulting roots can be directly used to find the general solution.

The **general solution of a given differential equation** is defined by the following formula:

\[ y( t ) \ = \ C_0 \ + \ C_1 e^ { r_1 t } \ + \ C_2 e^ { r_2 t } + \ … \ … \ … \ + \ C_n e^ { r_n t } \]

where $ y $ is the **dependent variable**, $ t $ is the **independent variable**, $ C_0, \ C_1, \ C_2, \ … \ … \ …, \ C_n $ are **constants of integration**, and $ r_0, \ r_1, \ r_2, \ … \ … \ …, \ r_n $ are the **roots of the polynomial**.

## Expert Answer

**Given:**

\[ y^{ ( 6 ) } \ – \ y^{ ( 2 ) } \ = \ 0 \]

Let **D be the differential operator**, then the above **equation reduces to**:

\[ D^{ 6 } \ – \ D^{ 2 } \ = \ 0 \]

\[ D^{ 2 } \bigg [ D^{ 4 } \ – \ 1 \bigg ] \ = \ 0 \]

\[ D^{ 2 } \bigg [ ( D^{ 2 } )^2 \ – \ ( 1 )^2 \bigg ] \ = \ 0 \]

\[ D^{ 2 } ( D^{ 2 } \ + \ 1 ) ( D^{ 2 } \ – \ 1 ) \ = \ 0 \]

\[ D^{ 2 } ( D^{ 2 } \ + \ 1 ) \bigg [ ( D )^2 \ – \ ( 1 )^2 \bigg ] \ = \ 0 \]

\[ D^{ 2 } ( D^{ 2 } \ + \ 1 ) ( D \ + \ 1 ) ( D \ – \ 1 ) \ = \ 0 \]

Hence the **roots of the equation** are:

\[ 0, \ 0, \ \pm 1, \pm i \]

According to the **general form** of the solution of a **differential equation**, for **our case**:

\[ y( t ) \ = \ \bigg ( C_0 \ + \ t C_1 \bigg ) e^ { ( 0 ) t } \ + \ C_2 e^ { ( +1 ) t } + \ C_3 e^ { ( -1 ) t } + \ C_4 cos ( t ) + \ C_5 sin ( t )Â \]

\[ y( t ) \ = \ C_0 \ + \ t C_1 \ + \ C_2 e^ { t } + \ C_3 e^ { -t } + \ C_4 cos ( t ) + \ C_5 sin ( t )Â \]

## Numerical Result

\[ y( t ) \ = \ C_0 \ + \ t C_1 \ + \ C_2 e^ { t } + \ C_3 e^ { -t } + \ C_4 cos ( t ) + \ C_5 sin ( t )Â \]

## Example

Given the equation $ y^{ ( 2 ) } \ – \ 1 \ = \ 0 $, **find a general solution**.

**The above equation reduces to:**

\[ ( D^{ 2 } \ – \ 1 \ = \ 0 \]

\[ \bigg [ ( D )^2 \ – \ ( 1 )^2 \bigg ] \ = \ 0 \]

\[ ( D \ + \ 1 ) ( D \ – \ 1 ) \ = \ 0 \]

So the **roots** are $ \pm 1 $ and the **general solution** is:

\[ y( t ) \ = \ C_0 \ + \ C_1 e^ { ( +1 ) t } \ + \ C_2 e^ { ( -1 ) t } \]