91Ó°ÊÓ

Consider the given differential equation,

$\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)-2\mathrm{x}\text{'}\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=24{\mathrm{t}}^2{\mathrm{e}}^{\mathrm{t}}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}…\left(1\right)$

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

$\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)-2\mathrm{x}\text{'}\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^2-2\mathrm{m}+1=0$

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2-2\mathrm{m}+1=0\\ {(\mathrm{m}-1)}^2=0\end{array}$

The roots of auxiliary equation are,

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

The complimentary solution of the given equation is;

${\mathrm{x}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{\mathrm{t}}$

Therefore, the particular solution of equation (1),

${\mathrm{x}}_{\mathrm{p}}={\mathrm{t}}^2({\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C}){\mathrm{e}}^{\mathrm{t}}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰}.....\left(2\right)$

Now find the derivative of above equation,

$\begin{array}{c}{\mathrm{x}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=({\mathrm{At}}^4+(\mathrm{B}+4\mathrm{A}){\mathrm{t}}^3+(\mathrm{C}+3\mathrm{B}){\mathrm{t}}^2+2\mathrm{Ct}){\mathrm{e}}^{\mathrm{t}}\\ {\mathrm{x}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=({\mathrm{At}}^4+(\mathrm{B}+8\mathrm{A}){\mathrm{t}}^3+(\mathrm{C}+6\mathrm{B}+12\mathrm{A}){\mathrm{t}}^2+(4\mathrm{C}+6\mathrm{B})\mathrm{t}+2\mathrm{C}){\mathrm{e}}^{\mathrm{t}}\end{array}$

From the equation (1), substitute the value of ${\mathbf{x}}_p\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{,}\text{ â¶Ä‰}{\mathbf{x}}_p\mathbf{\text{'}}\left(\mathbf{t}\right)$and ${\mathbf{x}}_p\left(\mathbf{t}\right)$, we get

$\begin{array}{l}⇒{\mathrm{x}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)-2{\mathrm{x}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)+{\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=24{\mathrm{t}}^2{\mathrm{e}}^{\mathrm{t}}\\ ⇒({\mathrm{At}}^4+(\mathrm{B}+8\mathrm{A}){\mathrm{t}}^3+(\mathrm{C}+6\mathrm{B}+12\mathrm{A}){\mathrm{t}}^2+(4\mathrm{C}+6\mathrm{B})\mathrm{t}+2\mathrm{C}){\mathrm{e}}^{\mathrm{t}}\\ \text{ â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰}-2({\mathrm{At}}^4+(\mathrm{B}+4\mathrm{A}){\mathrm{t}}^3+(\mathrm{C}+3\mathrm{B}){\mathrm{t}}^2+2\mathrm{Ct}){\mathrm{e}}^{\mathrm{t}}+{\mathrm{t}}^2({\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C}){\mathrm{e}}^{\mathrm{t}}=24{\mathrm{t}}^2{\mathrm{e}}^{\mathrm{t}}\\ ⇒(12{\mathrm{At}}^2+6\mathrm{Bt}+2\mathrm{C}){\mathrm{e}}^{\mathrm{t}}=24{\mathrm{t}}^2{\mathrm{e}}^{\mathrm{t}}\\ ⇒12{\mathrm{At}}^2{\mathrm{e}}^{\mathrm{t}}+6{\mathrm{Bte}}^{\mathrm{t}}+2{\mathrm{Ce}}^{\mathrm{t}}=24{\mathrm{t}}^2{\mathrm{e}}^{\mathrm{t}}\end{array}$

Comparing the all coefficients of the above equation,

$\begin{array}{c}12\mathrm{A}=24\text{ â¶Ä‰}⇒\mathrm{A}=2\\ \mathrm{B}=0\\ \mathrm{C}=0\end{array}$

Therefore, the particular solution of equation (1),

$\begin{array}{l}{\mathrm{x}}_{\mathrm{p}}={\mathrm{t}}^2({\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C}){\mathrm{e}}^{\mathrm{t}}\\ {\mathrm{x}}_{\mathrm{p}}={\mathrm{t}}^2(2{\mathrm{t}}^2+\left(0\right)\mathrm{t}+\left(0\right)){\mathrm{e}}^{\mathrm{t}}\\ {\mathrm{x}}_{\mathrm{p}}=2{\mathrm{t}}^4{\mathrm{e}}^{\mathrm{t}}\end{array}$