Subject : Applied Mathematics 2

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

Difficulty: High

Solution:

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

For complementary solution,

$\quad \boldsymbol{f}(\boldsymbol{D})=\mathbf{0}$

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

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

The complementary solution of given diff eqn is,

$\underset{\text { For complementary solution }}{\boldsymbol{y}_{c}=c_{1} \cos x+c_{2} \sin x+c_{3} e^{-x}}$

$\begin{aligned} & y_{p}=\frac{1}{f(D)} X=\frac{1}{\left(D^{3}+D^{2}+D+1\right)} \sin ^{2} x=\frac{1}{2\left(D^{3}+D^{2}+D+1\right)}(1-\cos 2 x) \\ &=\frac{1}{2\left(D^{3}+D^{2}+D+1\right)} e^{0 x}-\frac{1}{2\left(D^{3}+D^{2}+D+1\right)} \cos 2 x \\=& \frac{1}{2}-\frac{1}{2} \times \frac{1}{-2-4+D+1} \cos 2 x \\ &=\frac{1}{2}+\frac{1}{6} \cdot \frac{1}{D+1} \cos 2 x \end{aligned}$

$\begin{aligned} &=\frac{1}{2}+\frac{1}{6} \cdot \frac{1}{D+1} \frac{D-1}{D-1} \cos 2 x \\ &=\frac{1}{2}+\frac{1}{6} \cdot \frac{D-1}{D^{2}-1} \cos 2 x \\ &=\frac{1}{2}+\frac{1}{6} \cdot \frac{-2 \sin 2 x-\cos 2 x}{-4-1} \cos 2 x \\ y_{p} &=\frac{1}{2}+\frac{1}{30} \cdot(2 \sin 2 x+\cos 2 x) \end{aligned}$

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

$y_{g}=y_{c}+y_{p}=c_{1} \cos x+c_{2} \sin x+c_{3} e^{-x}+\frac{1}{2}+\frac{1}{30} \cdot(2 \sin 2 x+\cos 2 x)$