Question

Decide whether the method of undetermined coefficients together with superposition can be applied to find a particular solution of the given equation. Do not solve the equation. $\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}{\left({\mathit{e}}^{\mathit{t}}\mathbf{+}\mathit{t}\right)}^{\mathbf{2}}$

Short answer

Yes.

Step-by-step solution

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 non-negative 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.

Now, write the auxiliary equation of the above differential equation

The given differential equation is,

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

Write the homogeneous differential equation of equation (1),

$y\text{'}\text{'}-y\text{'}+y=0$

The auxiliary equation for the above equation,

$r^2-r+1=0$

Now find the roots of the auxiliary equation

Solve the auxiliary equation,

$$\begin{gathered} r^2-r+1=0 \\ r=\frac{1\pm \sqrt{1-4}}{2} \\ r=\frac{1\pm i\sqrt{3}}{2} \end{gathered}$$

The roots of the auxiliary equation are,

$r_1=\frac{1+i\sqrt{3}}{2}, r_2=\frac{1-i\sqrt{3}}{2}$

Final Conclusion

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,

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

The first condition is satisfied, one has:

M=0 and r = 2 are not a root of the associated auxiliary equation;

s = 0 if r is not a root of the associated auxiliary equation;

Therefore, the particular solution of the equation,

$$\begin{gathered} y_p\left(x\right)=t^s\left(A_m{t}^m+...+A_{1}t+A_0\right)e^{rt} \\ {y}_p\left(x\right)=t^0\left(A_0\right)e^{2t} \\ {y}_p\left(x\right)=A_0{e}^{2t} \end{gathered}$$

The second condition satisfied,

One has,

M=1 and r = 1 are not a root of the associated auxiliary equation;

s = 0 if r is not a root of the associated auxiliary equation;

Hence, the particular solution of the equation,

$$\begin{gathered} y_p\left(x\right)=t^s\left(A_m{t}^m+...+A_{1}t+A_0\right)e^{rt} \\ {y}_p\left(x\right)=t^0\left(A_{1}t+A_0\right)e^t \\ {y}_p\left(x\right)=\left(A_{1}t+A_0\right)e^t \end{gathered}$$

The third condition is satisfied, one has:

M=2 and r = 0 are not a root of the associated auxiliary equation;

s = 0 if r is not a root of the associated auxiliary equation;

Accordingly, the particular solution of the equation,

$$\begin{gathered} y_p\left(x\right)=t^s\left(A_m{t}^m+...+A_{1}t+A_0\right)e^{rt} \\ {y}_p\left(x\right)=t^0\left(A_2{t}^2+A_{1}t+A_0\right)e^{\left(0\right)t} \\ {y}_p\left(x\right)=A_2{t}^2+A_{1}t+A_0 \end{gathered}$$

R.H.S. of the equation $t^2,e^{2t}$and $2t{e}^t$is the combination of polynomials, exponentials, sines or cosines or product of these t function.

So, the method of undetermined coefficients can be applied.