Q38E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a particular solution to the given higher-order equation.

${\mathbf{y}}^4\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{10}\mathbf{cost}\mathbf{-}\mathbf{20}\mathbf{sint}$

Step-by-Step Solution

Answer

The particular solution is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{cost}-2\mathrm{sint}.$

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

Given equation,

 

${\mathrm{y}}^4-5\mathrm{y}\text{'}\text{'}+4\mathrm{y}=10\mathrm{cost}-20\mathrm{sint}\mathrm{......}\left(\mathbf{1}\right)$

 

Consider the particular solution is,

 

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acost}+\mathrm{Bsint}. .....\left(2\right)$

 

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=-\mathrm{Asint}+\mathrm{Bcost}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=-\mathrm{Acost}-\mathrm{Bsint}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)=\mathrm{Asint}-\mathrm{Bcost}\\ {{\mathrm{y}}_{\mathrm{p}}}^4\left(\mathrm{t}\right)=\mathrm{Acost}+\mathrm{Bsint}\end{array}$

Substitute value of ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right),{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$  and ${{\mathrm{y}}_{\mathrm{p}}}^4\left(\mathrm{t}\right)$ in the equation (1),

$\begin{array}{c}{\mathrm{y}}^4-5\mathrm{y}\text{'}\text{'}+4\mathrm{y}=10\mathrm{cost}-20\mathrm{sint}\\ \mathrm{Acost}+\mathrm{Bsint}-5(-\mathrm{Acost}-\mathrm{Bsint})+4(\mathrm{Acost}+\mathrm{Bsint})=10\mathrm{cost}-20\mathrm{sint}\\ 10\mathrm{Acost}+10\mathrm{Bsint}=10\mathrm{cost}-20\mathrm{sint}\end{array}$

Comparing the coefficients of the above equation;

$\begin{array}{c}10\mathrm{A}=10\Rightarrow \mathrm{A}=1\\ 10\mathrm{B}=-20\Rightarrow \mathrm{B}=-2\end{array}$

Step 2: Final conclusion.

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

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acost}+\mathrm{Bsint}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{cost}-2\mathrm{sint}\end{array}$