Subject : Applied Mathematics 2

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

Difficulty: High

Solution:

$\begin{aligned} & \frac{d}{d x}=\mathrm{D} \\ & \therefore D^{4} y+2 D^{2} y+y=0 \\ & \therefore D^{4}+2 D^{2}+1=0 \\ \text { Put } & D^{2}=t \end{aligned}$

$\begin{aligned} & \text { Put } D^{2}=t \\ \Rightarrow & t^{2}+2 t+1=0 \\=& t=-1,-1 \end{aligned}$

Roots are: $\quad \mathrm{D}=+\mathrm{i}, -\mathrm{i},+\mathrm{i},-\mathrm{i}$

The complementary solution of given eqn is

$\quad y_{c}=y_{g}=\left(C_{1}+x C_{2}\right) \cos x+\left(C_{3}+x C_{4}\right) \sin x$