Q35E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 35, use the method of undetermined coefficients to find a particular solution to the given higher-order equation.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}{\mathbf{te}}^t$

Step-by-Step Solution

Answer

The particular solution is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{t}\left(\frac{1}{10}\mathrm{t}-\frac{4}{25}\right){\mathrm{e}}^{\mathrm{t}}.$

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

The given differential equation is:

 

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

 

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

 

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

 

The auxiliary equation for the above equation,

 ${\mathrm{m}}^3+{\mathrm{m}}^2-2=0$

 Solve the auxiliary equation,

$\begin{array}{c}(\mathrm{m}-1)({\mathrm{m}}^2+2\mathrm{m}+2)=0\\ \mathrm{m}=1,\mathrm{m}=\frac{-2\pm \sqrt{4-8}}{2}\\ \mathrm{m}=1,\mathrm{m}=-1\pm \mathrm{i}\end{array}$

Step 2: Use the method of undetermined coefficients to find a particular solution to a given differential equation.

Consider the particular solution is,

 

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

 

Take the first, second, and third derivative of the above equation,

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

Substitute value of ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right),{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$ and ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)$ in the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}\text{'}+\mathrm{y}\text{'}\text{'}-2\mathrm{y}={\mathrm{te}}^{\mathrm{t}}\\ ({\mathrm{At}}^2+(6\mathrm{A}+\mathrm{B})\mathrm{t}+(6\mathrm{A}+3\mathrm{B})){\mathrm{e}}^{\mathrm{t}}+({\mathrm{At}}^2+(4\mathrm{A}+\mathrm{B})\mathrm{t}+(2\mathrm{A}+2\mathrm{B})){\mathrm{e}}^{\mathrm{t}}-2({\mathrm{At}}^2+\mathrm{Bt}){\mathrm{e}}^{\mathrm{t}}={\mathrm{te}}^{\mathrm{t}}\\ [10\mathrm{At}+(8\mathrm{A}+5\mathrm{B})]{\mathrm{e}}^{\mathrm{t}}={\mathrm{te}}^{\mathrm{t}}\end{array}$

 

Comparing the coefficients of the above equation;

  $\begin{array}{c}10\mathrm{A}=1\\ \mathrm{A}=\frac{1}{10}\\ 8\mathrm{A}+5\mathrm{B}=0...\left(3\right)\end{array}$

Substitute the value A in the equation (3),

$\begin{array}{c}8\left(\frac{1}{10}\right)+5\mathrm{B}=0\\ \frac{4}{5}+5\mathrm{B}=0\\ \mathrm{B}=-\frac{4}{25}\end{array}$

Step 3: Conclusion.

Substitute values A and B in the equation (2),

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{t}(\mathrm{At}+\mathrm{B}){\mathrm{e}}^{\mathrm{t}}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{t}\left(\frac{1}{10}\mathrm{t}-\frac{4}{25}\right){\mathrm{e}}^{\mathrm{t}}\end{array}$

 

Therefore, the particular solution of the equation (1),

 

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{t}\left(\frac{1}{10}\mathrm{t}-\frac{4}{25}\right){\mathrm{e}}^{\mathrm{t}}$