Q21E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a particular solution to the differential equation.

$\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{+}\mathbf{4}\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{te}}^{2t}$

Step-by-Step Solution

Answer

The particular solution is  ${\mathbf{x}}_p\mathbf{=}\frac{1}{6}{\mathbf{t}}^3{\mathbf{e}}^{2t}.$

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

Consider the given differential equation,

 

$\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)-4\mathrm{x}\text{'}\left(\mathrm{t}\right)+4\mathrm{x}\left(\mathrm{t}\right)={\mathrm{te}}^{2\mathrm{t}}\dots \left(1\right)$

 

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

 

$\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)-4\mathrm{x}\text{'}\left(\mathrm{t}\right)+4\mathrm{x}\left(\mathrm{t}\right)=0$

 

The auxiliary equation for the above equation,

 

${\mathrm{m}}^2-4\mathrm{m}+4=0$

Step 2: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

 

$\begin{array}{c}{\mathrm{m}}^2-4\mathrm{m}+4=0\\ {(\mathrm{m}-2)}^2=0\end{array}$

 

The roots of the auxiliary equation are;

 

${\mathrm{m}}_1=2,\&{\mathrm{m}}_2=2$

 

The complementary solution of the given equation is;

 

${\mathrm{x}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{2\mathrm{t}}$

 

Step 2: Find a particular solution.

Therefore, the particular solution of equation (1),

${\mathrm{x}}_{\mathrm{p}}={\mathrm{t}}^2(\mathrm{At}+\mathrm{B}){\mathrm{e}}^{2\mathrm{t}}\dots \left(2\right)$

Now find the derivative of the above equation,

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

 

From the equation (1), substitute the value of  ${\mathrm{x}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right),{\mathrm{x}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)$ and ${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)$, we get

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

Step 4: Final conclusion.

Comparing all coefficients of the above equation,

 $\begin{array}{c}6\mathrm{A}=1\Rightarrow \mathrm{A}=\frac{1}{6}\\ \mathrm{B}=0\end{array}$

 

Therefore, the particular solution of equation (1),

 $\begin{array}{c}{\mathrm{x}}_{\mathrm{p}}={\mathrm{t}}^2(\mathrm{At}+\mathrm{B}){\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{x}}_{\mathrm{p}}={\mathrm{t}}^2(\frac{}{}\mathrm{t}+0){\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{x}}_{\mathrm{p}}=\frac{}{}{\mathrm{t}}^3{\mathrm{e}}^{2\mathrm{t}}\end{array}$