Q22E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a particular solution to the differential equation.

$\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{+}\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\mathbf{24}{\mathbf{t}}^2{\mathbf{e}}^t$

Step-by-Step Solution

Answer

The particular solution of the differential equation  ${\mathbf{x}}_p\mathbf{=}\mathbf{2}{\mathbf{t}}^4{\mathbf{e}}^t.$

Step 1: Firstly, write the auxiliary equation of the above differential equation.

Consider the given differential equation,

 

$\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)-2\mathrm{x}\text{'}\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=24{\mathrm{t}}^2{\mathrm{e}}^{\mathrm{t}}\dots \left(1\right)$

 

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,

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

Step 2: Now find the roots of auxiliary equation

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, 

 

${\mathrm{m}}_1=1,\&{\mathrm{m}}_2=1$

 

The complimentary solution of the given equation is;

 

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

Step 3: Use the method of undetermined coefficients to find a particular solution to the differential equation.

Therefore, the particular solution of equation (1),

 ${\mathrm{x}}_{\mathrm{p}}={\mathrm{t}}^2({\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C}){\mathrm{e}}^{\mathrm{t}}.....\left(2\right)$

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  ${\mathbf{x}}_p\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{,}{\mathbf{x}}_p\mathbf{\text{'}}\left(\mathbf{t}\right)$ and ${\mathbf{x}}_p\left(\mathbf{t}\right)$, we get

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

Step 4: Final conclusion.

Comparing the all coefficients of the above equation,

 

$\begin{array}{c}12\mathrm{A}=24\Rightarrow \mathrm{A}=2\\ \mathrm{B}=0\\ \mathrm{C}=0\end{array}$

 

Therefore, the particular solution of equation (1),

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