Q40E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a particular solution to the given higher-order equation.${\mathbf{y}}^4\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{6}\mathbf{t}\mathbf{-}\mathbf{20}$

Step-by-Step Solution

Answer

The particular solution to the given higher-order equation is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=-3{\mathrm{t}}^2+2\mathrm{t}.$

Step 1: Consider the particular solution for the given differential equation.

The given differential equation is,

 

${\mathrm{y}}^4-3\mathrm{y}\text{'}\text{'}\text{'}+3\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}=6\mathrm{t}-20. .....\left(1\right)$

 

Consider the particular solution is,

 

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

 

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

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=2\mathrm{At}+\mathrm{B}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=2\mathrm{A}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)=0\\ {{\mathrm{y}}_{\mathrm{p}}}^4\left(\mathrm{t}\right)=0\end{array}$

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

$\begin{array}{c}{\mathrm{y}}^4-3\mathrm{y}\text{'}\text{'}\text{'}+3\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}=6\mathrm{t}-20\\ 0-3\left(0\right)+3\left(2\mathrm{A}\right)-(2\mathrm{At}+\mathrm{B})=6\mathrm{t}-20\\ -2\mathrm{At}+(-\mathrm{B}+6\mathrm{A})=6\mathrm{t}-20\end{array}$

Comparing all coefficients of the above equation;

 $\begin{array}{c}-2\mathrm{A}=6\Rightarrow \mathrm{A}=-3\\ -\mathrm{B}+6\mathrm{A}=-20. .....\left(3\right)\end{array}$

Substitute the value A in the equation (3),

$\begin{array}{c}-\mathrm{B}+6(-3)=-20\\ \mathrm{B}=2\end{array}$

Step 2: Final conclusion.

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

 $\begin{array}{l}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{At}}^2+\mathrm{Bt}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=-3{\mathrm{t}}^2+2\mathrm{t}\end{array}$