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Q10E

Expert-verifiedFound in: Page 186

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**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. ${\mathit{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathit{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathit{y}}{\mathbf{=}}{\left({\mathit{e}}^{\mathit{t}}\mathbf{+}\mathit{t}\right)}^{{\mathbf{2}}}$**

Yes.

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

${y}_{p}\left(x\right)={t}^{s}\left({A}_{m}{t}^{m}+...+{A}_{1}t+{A}_{0}\right){e}^{rt}$

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

The given differential equation is,

$y\text{'}\text{'}-y\text{'}+y={\left({e}^{t}+t\right)}^{2}\phantom{\rule{0ex}{0ex}}y\text{'}\text{'}-y\text{'}+y={e}^{2t}+{t}^{2}+2t{e}^{t}\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a...\left(1\right)$

Write the homogeneous differential equation of equation (1),

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

The auxiliary equation for the above equation,

${r}^{2}-r+1=0$

Solve the auxiliary equation,

${r}^{2}-r+1=0\phantom{\rule{0ex}{0ex}}r=\frac{1\pm \sqrt{1-4}}{2}\phantom{\rule{0ex}{0ex}}r=\frac{1\pm i\sqrt{3}}{2}$

The roots of the auxiliary equation are,

${r}_{1}=\frac{1+i\sqrt{3}}{2},\u200a\u200a\u200a\u200a\u200a\u200a{r}_{2}=\frac{1-i\sqrt{3}}{2}$

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,

${y}_{p}\left(x\right)={t}^{s}\left({A}_{m}{t}^{m}+...+{A}_{1}t+{A}_{0}\right){e}^{rt}\phantom{\rule{0ex}{0ex}}{y}_{p}\left(x\right)={t}^{0}\left({A}_{0}\right){e}^{2t}\phantom{\rule{0ex}{0ex}}{y}_{p}\left(x\right)={A}_{0}{e}^{2t}$

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,

${y}_{p}\left(x\right)={t}^{s}\left({A}_{m}{t}^{m}+...+{A}_{1}t+{A}_{0}\right){e}^{rt}\phantom{\rule{0ex}{0ex}}{y}_{p}\left(x\right)={t}^{0}\left({A}_{1}t+{A}_{0}\right){e}^{t}\phantom{\rule{0ex}{0ex}}{y}_{p}\left(x\right)=\left({A}_{1}t+{A}_{0}\right){e}^{t}$

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,

${y}_{p}\left(x\right)={t}^{s}\left({A}_{m}{t}^{m}+...+{A}_{1}t+{A}_{0}\right){e}^{rt}\phantom{\rule{0ex}{0ex}}{y}_{p}\left(x\right)={t}^{0}\left({A}_{2}{t}^{2}+{A}_{1}t+{A}_{0}\right){e}^{\left(0\right)t}\phantom{\rule{0ex}{0ex}}{y}_{p}\left(x\right)={A}_{2}{t}^{2}+{A}_{1}t+{A}_{0}$

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.

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