Q36E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}{\mathit{t}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{2}\mathit{t}}\mathbf{-}{\mathit{e}}^{\mathbf{2}\mathit{t}}$.

Step-by-Step Solution

Answer

Thus, the answer is:

$y_p\left(x\right)=\left(A_2{t}^4+A_1{t}^3+A_0{t}^2\right)e^{2t}$

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

The differential equation is,

 $y\text{'}\text{'}-4y\text{'}+4y=t^2{e}^{2t}-e^{2t} ...\left(1\right)$

Write the homogeneous differential equation of equation (1),

$y\text{'}\text{'}-4y\text{'}+4y=0$ 

The auxiliary equation for the above equation,

$m^2-4m+4=0$

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

Solve the auxiliary equation,

$$\begin{gathered} m^2-4m+4=0 \\ \left(m-2\right)\left(m-2\right)=0 \end{gathered}$$ 

The roots of the auxiliary equation are, 

$m_1=2, m_2=2$ 

The complimentary solution of the given equation is,

$y_c=c_1{e}^{2t}+c_{2}t{e}^{2t}$

Step 3: Use the method of undetermined coefficients

The given differential equation is in the form of 

$ax\text{'}\text{'}+bx\text{'}+cx=e^{rt}$ 

According to the method of undetermined coefficients, 

 

To find a particular solution to the differential equation

$ay\text{'}\text{'}\left(x\right)+by\text{'}\left(x\right)+cy\left(x\right)=C{t}^m{e}^{rt}$ 

Where m is a nonnegative integer, use the form

$y_p\left(x\right)=t^s\left(A_m{t}^m+...+A_{1}t+A_0\right)e^{rt}$

1. s = 0 if r is not a root of the associated auxiliary equation; 
2. s = 1 if r is a simple root of the associated auxiliary equation; 
3. s = 2 if r is a double root of the associated auxiliary equation

Step 4: Now find the form of a particular solution

To find a particular solution to the differential equation

 $ay\text{'}\text{'}\left(x\right)+by\text{'}\left(x\right)+cy\left(x\right)=C{t}^m{e}^{rt}$

Compare with the given differential equation,

$y\text{'}\text{'}-4y\text{'}+4y=t^2{e}^{2t}-e^{2t}$ 

Condition satisfied, 

 

M=2, s = 2 if r = 2 is a double root of the associated auxiliary equation.

 

Therefore, the particular solution of the equation,

$$\begin{gathered} y_p\left(x\right)=t^2\left(A_2{t}^2+A_{1}t+A_0\right)e^{2t} \\ {y}_p\left(x\right)=\left(A_2{t}^4+A_1{t}^3+A_0{t}^2\right)e^{2t} \end{gathered}$$