Q35E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{5}\mathbf{t}}\mathbf{+}\mathit{t}\mathit{s}\mathit{i}\mathit{n}\mathbf{3}\mathit{t}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{3}\mathit{t}$

Step-by-Step Solution

Answer

Thus, the answer is: $y_p\left(t\right)=\left(A_{1}t+A_0\right)cos3x+\left(B_{1}t+B_0\right)sin3x+C{e}^{5t}$

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

The differential equation is,

 $y\text{'}\text{'}-4y\text{'}+5y=e^{5t}+tsin3t-cos3t ...\left(1\right)$

Write the homogeneous differential equation of equation (1),

 $y\text{'}\text{'}-4y\text{'}+5y=0$

 

The auxiliary equation for the above equation,

$m^2-4m+5=0$ 

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

Solve the auxiliary equation,

$$\begin{gathered} m^2-4m+5=0 \\ m=\frac{4\pm \sqrt{16-20}}{2} \\ m=\frac{4\pm \sqrt{-4}}{2} \\ m=2\pm i \end{gathered}$$ 

The roots of the auxiliary equation are, 

 $m_1=2+i, m_2=2-i$

Step 3: Use the method of undetermined coefficients

According to the method of undetermined coefficients, 

 

To find a particular solution to the differential equation 

 $$\begin{gathered} ay\text{'}\text{'}+by\text{'}+cy=\left\{C{t}^m{e}^{\alpha t}cos\beta t \\ or \\ C{t}^m{e}^{\alpha t}sin\beta t\right\} ...\left(2\right) \end{gathered}$$

For $\beta \ne 0$, use the form

$y_p\left(x\right)=t^s\left[\left(A_m{t}^m+...+A_{1}t+A_0\right)e^{\alpha t}cos\beta t+t^s\left(B_m{t}^m+...+B_{1}t+B_0\right)e^{\alpha t}sin\beta t\right]$ 

With s = 1 if $\alpha +i\beta$ is a root of the associated auxiliary equation.

And s = 0 if $\alpha +i\beta$ is not a root of the associated auxiliary equation.

Step 4: Now find the form of a particular solution

Compare equations (1) and (2),

 

One gets, M=1 and $\alpha =0;\beta =3$

 

Therefore, $\alpha +i\beta =3$ is not a root of the associated auxiliary equation.

 

So, s = 0.

 

Assume, the particular solution of equation (1),

$$\begin{gathered} y_p\left(t\right)=t^s\left(A_m{t}^m+...+A_{1}t+A_0\right)e^{\alpha x}cos\beta x+t^s\left(B_m{t}^m+...+B_{1}t+B_0\right)e^{\alpha x}sin\beta x+C{e}^{5t} \\ {y}_p\left(t\right)=t^0\left(A_{1}t+A_0\right)e^{\left(0\right)x}cos\left(3\right)x+t^0\left(B_{1}t+B_0\right)e^{\left(0\right)x}sin\left(3\right)x+C{e}^{5t} \\ {y}_p\left(t\right)=\left(A_{1}t+A_0\right)cos3x+\left(B_{1}t+B_0\right)sin3x+C{e}^{5t} \end{gathered}$$