The aim of this problem is to understand the general solution to the higher order differential equations. 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:
\[ y( t ) \ = \ \bigg ( C_0 \ + \ t C_1 \bigg ) e^ { ( 0 ) t } \ + \ C_2 e^ { ( +1 ) t } + \ C_3 e^ { ( -1 ) t } + \ C_4 cos ( t ) + \ C_5 sin ( t ) \]
\[ y( t ) \ = \ C_0 \ + \ t C_1 \ + \ C_2 e^ { t } + \ C_3 e^ { -t } + \ C_4 cos ( t ) + \ C_5 sin ( t ) \]
Numerical Result
\[ y( t ) \ = \ C_0 \ + \ t C_1 \ + \ C_2 e^ { t } + \ C_3 e^ { -t } + \ C_4 cos ( t ) + \ C_5 sin ( t ) \]
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:
\[ y( t ) \ = \ C_0 \ + \ C_1 e^ { ( +1 ) t } \ + \ C_2 e^ { ( -1 ) t } \]