Question

1.Solve the differential equation (2𝐷² + 5𝐷 + 2)𝑦 = 𝑒^{-1/ 2 x}

2. Solve the differential equation (𝐷3 − 3𝐷2 + 4𝐷 − 2)𝑦 = 𝑐𝑜𝑠𝗑.

Answer 1

1. The auxiliary equation is, $2m^2 + 5m+2=0$.

Solving the equation we get, $m = -\frac{1}{2}, -2$

The complementary function is $\text{C.F} = c_1 e^{-\frac{x}{2}} + c_2 e^{-2x}$

$\begin{aligned} \text{Particular Integral} &= \frac{1}{2D^2+5D+2}e^{-\frac{x}{2}}\\ &= \frac{1}{2(-\frac{1}{2})^2+5(-\frac{1}{2})+2}e^{-\frac{x}{2}}~~(\text{Replace D by }-\frac{1}{2})\\ &=x \frac{1}{4D+5}e^{-\frac{x}{2}}~~~ (\text{Rule failed in the above step})\\ &=\frac{x e^{-\frac{x}{2}}}{3}~~(\text{Replace D by }-\frac{1}{2})\\ \end{aligned}$

The general solution is $y = c_1 e^{-\frac{x}{2}} + c_{2}e^{-2x} + \dfrac{xe^{-\frac{x}{2}}}{3}$

b. The auxiliary equation is, $m^3-3m^2+4m-2=0.$ Using synthetic division,

$\begin{array}{c|ccc} 1&1 &-3 & 4 & -2\\ & 0 & ~~1 & -2 & ~~~2\\ \hline &1 & ~-2 & ~~2 & \mid~0\\ \hline \end{array}$

$\therefore m^3-3m^2+4m-2 = (m-1)(m^2-2m+2) = 0 \implies m = 1, \\ m^2-2m+2 =0.\\$

Solving the quadratic,

$m = \dfrac{2\pm\sqrt{2^2-4\times1\times2}}{2} = \dfrac{2\pm2i}{2} = 1\pm i$ .

Hence the complementary function is,

$\text{C.F} = c_1e^{x} + e^{x}(c_2 \cos x + c_3 \sin x)$

$\begin{aligned} \text{Particular Integral} &= \frac{1}{D^3 - 3D^2+4D-2}\cos x \\ &= \frac{1}{-D + 3 + 4D - 2}\cos x ~~(\text{Replacing $D^2$ by $-1$})\\ &= \frac{1}{1 + 3D}\cos x\\ &= \frac{1-3D}{1 - 9D^{2}}\cos x~~~~~~~~~~~(\text{Multiplying the conjugate})\\ &= \frac{\cos x - 3D(\cos x)}{10}~~~~~(\text{Replacing $D^2$ by $-1$})\\ &=\frac{\cos x + 3\sin x}{10}\\ \end{aligned}$

The general solution is,

$y = e^x(c_1 + c_{2} \cos x + c_3 \sin x) + \dfrac{\cos x + 3\sin x}{10}$