Subject : Applied Mathematics 2

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

Difficulty: High

Solution:

$f(D)=0$

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

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

$D-1=0 \space $ for two times $\quad \left(D^{2}+1\right)=0 \space $ for two times

$\therefore d - 1 = 0 \quad \quad \quad \therefore D^2 = -1 $

Roots are: D = 1, 1, +i, +i, -i, -i

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