91Ó°ÊÓ

According to the method of undetermined coefficients, to find a particular solution to the differential equation,

$\mathrm{ay}\text{'}\text{'}+\mathrm{by}\text{'}+\mathrm{cy}=\left\{\begin{array}{l}{\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{³¦´Ç²õβ³Ù}\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}\mathrm{or}\\ {\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{²õ¾\pm ²Ôβ³Ù}\end{array}\right\}$

For,$\mathrm{β}≠0$ then use the form;

$({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{³¦´Ç²õβ³Ù}+{\mathrm{t}}^{\mathrm{s}}({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{²õ¾\pm ²Ôβ³Ù}$

With s = 1 if$\mathrm{Î\pm }+\mathrm{¾\pm β}$ is a root of the associated auxiliary equation, and s = 0 if$\mathrm{Î\pm }+\mathrm{¾\pm β}$ is not a root of the associated auxiliary equation.

The differential equation is,

$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+2\mathrm{y}=8{\mathrm{t}}^3{\mathrm{e}}^{-\mathrm{t}}\mathrm{sint}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰}…\left(1\right)$

Write the homogeneous differential equation of the equation (1),

$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+2\mathrm{y}=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^2+2\mathrm{m}+2=0$

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2+2\mathrm{m}+2=0\\ \mathrm{m}=\frac{-2Â\pm \sqrt{2^2-4\left(1\right)\left(2\right)}}{2\left(1\right)}\\ \mathrm{m}=\frac{-2Â\pm \sqrt{-4}}{2}\\ \mathrm{m}=-1Â\pm \mathrm{i}\end{array}$

The roots of the auxiliary equation are,

${\mathrm{m}}_1=-1+\mathrm{i},\text{ â¶Ä‰â¶Ä‰}\&\text{ â¶Ä‰â¶Ä‰}{\mathrm{m}}_2=-1-\mathrm{i}$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{e}}^{-\mathrm{x}}({\mathrm{c}}_1\mathrm{cosx}+{\mathrm{c}}_2\mathrm{sinx})$

To find a particular solution to the differential equation

$\mathrm{ay}\text{'}\text{'}+\mathrm{by}\text{'}+\mathrm{cy}=\left\{\begin{array}{l}{\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{³¦´Ç²õβ³Ù}\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}\mathrm{or}\\ {\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{²õ¾\pm ²Ôβ³Ù}\end{array}\right\}$

Compare with the given differential equation,

$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+2\mathrm{y}=8{\mathrm{t}}^3{\mathrm{e}}^{-\mathrm{t}}\mathrm{sint}$

We have,

$\mathrm{m}=3,\text{ }\mathrm{Î\pm }=-1,\text{ â¶Ä‰}\mathrm{β}=1$

And

$\mathrm{Î\pm }+\mathrm{¾\pm β}=-1+\mathrm{i}={\mathrm{m}}_1$

Therefore, we get

s = 1 if$\mathrm{Î\pm }+\mathrm{¾\pm β}$ is a root of the associated auxiliary equation.

The particular solution to the differential equation for m = 3,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{t}}^{\mathrm{s}}[({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{³¦´Ç²õβ³Ù}+{\mathrm{t}}^{\mathrm{s}}({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{²õ¾\pm ²Ôβ³Ù}]\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{t}[({\mathrm{A}}_3{\mathrm{t}}^3+{\mathrm{A}}_2{\mathrm{t}}^2+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{-\mathrm{t}}\mathrm{cost}+({\mathrm{B}}_3{\mathrm{t}}^3+{\mathrm{B}}_2{\mathrm{t}}^2+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{-\mathrm{t}}\mathrm{sint}]\end{array}$