Find the general solution of y^((7)) + 3y^((6)) + 3y^((5)) +…

Find the general solution of $y^{(7)} + 3y^{(6)} + 3y^{(5)} + y^{(4)} = 0$.

Step 1: Characteristic equation

Substituting $y=e^{rx}$: $r^7 + 3r^6 + 3r^5 + r^4 = 0.$

Step 2: Factor

Factor out $r^4$; the remaining cubic is a perfect cube, since $(r+1)^3=r^3+3r^2+3r+1$: $r^4(r^3+3r^2+3r+1)=0 \implies r^4(r+1)^3=0.$

Step 3: Roots

\[ r_1=0 \ \ (\text{multiplicity } 4), \qquad r_2=-1 \ \ (\text{multiplicity } 3). \]

Step 4: Fundamental system

A real root $r$ of multiplicity $m$ contributes $e^{rx},xe^{rx},\dots,x^{m-1}e^{rx}$: \[y_1=1,\ y_2=x,\ y_3=x^2,\ y_4=x^3, \qquad y_5=e^{-x},\ y_6=xe^{-x},\ y_7=x^2e^{-x}.\]

Step 5: Wronskian

Writing the equation as $y^{(7)}+p_6y^{(6)}+\dots=0$, the coefficient of $y^{(6)}$ is $p_6=3$, so by Abel’s theorem: \[W' = -p_6W = -3W \implies W = W_0e^{-3x} \neq 0.\]

Step 6: General solution

The solution is recorded in the result below.