$(D^2 – 3D + 2) y = 2e^x sin (\frac{x}{2})$

Subject : Applied Mathematics 2

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

Difficulty: High

Solution:

$\left(D^{2}-3 D+2\right) y=2 e^{x} \sin \left(\frac{x}{2}\right)$

For complementary function,

$\begin{aligned} f(D) &=0 \\ \therefore\left(D^{2}-3 D+2\right) &=0 \end{aligned}$

Roots are: D = 2,1 $\quad$ Real roots,

$ \mathbf{ y_{c}=c_{1} e^{x}+c_{2} e^{2 x} }$

For particular integral,

$\begin{aligned} y_{p} &=\frac{1}{f(D)} X \\ &=\frac{1}{\left(D^{2}-3 D+2\right)} 2 e^{x} \sin \left(\frac{x}{2}\right) \\ &=2 e^{x} \frac{1}{(D+1)^{2}-3(D+1)+2} \sin \left(\frac{x}{2}\right) \\ &=2 e^{x} \frac{1}{\left(D^{2}-D\right)} \sin \left(\frac{x}{2}\right) \\ &=2 e^{x} \frac{1}{-\left(\frac{1}{4}\right)-D} \sin \left(\frac{x}{2}\right) \\ &=-8 e^{x} \frac{1}{4 D+1} \sin \left(\frac{x}{2}\right) \\ &=-8 e^{x} \frac{4 D-1}{16 D^{2}-1} \sin \left(\frac{x}{2}\right) \\ \therefore y_{p} &= \frac{8}{5} e^{x}\left(-\sin \left(\frac{x}{2}\right)-2 \cos \left(\frac{x}{2}\right)\right) \end{aligned}$

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

$\mathbf{ \quad y_{c}=y_{c}+y_{p}=c_{1} e^{x}+c_{2} e^{2 x}+\frac{8}{5} e^{x}\left(-\sin \left(\frac{x}{2}\right)-2 \cos \left(\frac{x}{2}\right)\right) }$