Q34E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine the form of a particular solution for the differential equation. Do not solve

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{t}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{t}$

Step-by-Step Solution

Answer

Thus, the answer is: $y_p=Asint+Bcost+Ccos2t+Dsin2t$

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

The differential equation is,

 $y\text{'}\text{'}+5y\text{'}+6y=sint-cos2t ...\left(1\right)$

Write the homogeneous differential equation of equation (1),

$y\text{'}\text{'}+5y\text{'}+6y=0$ 

The auxiliary equation for the above equation,

$m^2+5m+6=0$

Step 2: Now find the complementary solution of the given equation is,

Solve the auxiliary equation,

$$\begin{gathered} m^2+5m+6=0 \\ m^2+3m+2m+6=0 \\ m\left(m+3\right)+2\left(m+3\right)=0 \\ \left(m+2\right)\left(m+3\right)=0 \end{gathered}$$ 

The roots of the auxiliary equation are, 

 $m_1=-2, m_2=-3$

The complementary solution of the given equation is,

$y_c=c_1{e}^{-2t}+c_2{e}^{-3t}$

Step 3: Now find the form of a particular solution

The particular solution to equation (1) will be the linear combination of terms $sint, cos2t$ and their independent derivatives.

 

Therefore, the particular solution of equation (1),

$y_p=Asint+Bcost+Ccos2t+Dsin2t$