$(D^3 + 1)^2 y = 0$

Subject : Applied Mathematics 2

Topic : Linear Differential Equations With Constant Coefficients and Variable Coefficients Of Higher Order

Difficulty: High

Solution:

$\left(D^{3}+1\right)^{2} y=0$

For complementary solution,

$f(D)=0$

$\therefore\left(D^{3}+1\right)^{2}=0$

$\therefore\left(D^{3}+1\right)=0$

Roots are $: D=-1, \frac{1}{2}+i \frac{\sqrt{3}}{2}, \frac{1}{2}-i \frac{\sqrt{3}}{2} \quad . .$ for two times

Roots of given diff. eqn are real and complex

The general solution of given diff. eqn is given by,

$\quad y_{g}=y_{c}=\left(c_{1}+x c_{2}\right) e^{-x}+e^{\frac{x}{2}}\left[\left(c_{3}+x c_{4}\right) \cos x+\left(c_{5}+x c_{6}\right) \sin x\right]$