Q36E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 36, use the method of undetermined coefficients to find a particular solution to the given higher-order equation. ${\mathit{y}}^{\mathbf{4}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{t}$

Step-by-Step Solution

Answer

The value is $y_p\left(t\right)=-\frac{1}{4}\sin ty$

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

The given differential equation is:

 $y^4-3y\text{'}\text{'}-8y=\sin t ...\left(\mathbf{1}\right)$

 

Write the homogeneous differential equation of equation (1),

 $y^4-3y\text{'}\text{'}-8y=0$

 

The auxiliary equation for the above equation,

$m^4-3m^2-8=0$ 

 

Solve the auxiliary equation,

$m^4-3m^2-8=0$ 

Let $m^2=t$,

$$\begin{gathered} t^2-3t-8=0 \\ t=\frac{3\pm \sqrt{9+32}}{2} \\ t=\frac{3\pm \sqrt{41}}{2} \end{gathered}$$

 $m={\left(\frac{3\pm \sqrt{41}}{2}\right)}^2$

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

According to the method of undetermined coefficients, consider the particular solution is,

$y_p\left(t\right)=A\sin t+B\cos t ...\left(\mathbf{2}\right)$ 

 

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

$$\begin{gathered} y_p\text{'}\left(t\right)=A\cos t-B\sin t \\ y_p\text{'}\text{'}\left(t\right)=-A\sin t-B\cos t \\ {y}_p\text{'}\text{'}\text{'}\left(t\right)=-A\cos t+B\sin t \\ { y_p}^4\left(t\right)=A\sin t+B\cos t \end{gathered}$$ 

 

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

$$\begin{gathered} y^4-3y\text{'}\text{'}-8y=\sin t \\ A\sin t+B\cos t-3\left(-A\sin t-B\cos t\right)-8\left(A\sin t+B\cos t\right)=\sin t \\ -4A\sin t-4B\cos t=\sin t \end{gathered}$$ 

Comparing the coefficients of the above equation;

 $$\begin{gathered} -4A=1 \\ A=-\frac{1}{4} \\ B=0 \end{gathered}$$

Step 3: Conclusion.

Substitute values A and B in equation (2).

 

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

$$\begin{gathered} y_p\left(t\right)=A\sin t+B\cos t \\ {y}_p\left(t\right)=-\frac{1}{4}\sin t+\left(0\right)\cos t \\ {y}_p\left(t\right)=-\frac{1}{4}\sin t \end{gathered}$$