Question

Consider the differential equation $\left(D-a\right)\left(D-b\right)\mathbf{y}\mathbf{=}{\mathbf{P}}_{\mathbf{n}}\left(x\right)$, where ${\mathbf{P}}_{\mathbf{n}}\left(X\right)$is a polynomial of degree $\mathbf{n}$. Show that a particular solution of this equation is given by $\left(6.24\right)$with $\mathbf{c}\mathbf{=}\mathbf{0}$; that is, ${\mathbf{y}}_{\mathbf{p}}$is $\{\begin{array}{ll}a\mathrm{polynomial}{Q}_n\left(x\right)\mathrm{of}\mathrm{degree}n\mathrm{if}a\mathrm{and}b\mathrm{are}\mathrm{both}\mathrm{different}\mathrm{from}\mathrm{zero};& \\ {\mathrm{xQ}}_n\left(x\right)\mathrm{if}a\not\equiv 0,\mathrm{but}b=0& \\ x^2{Q}_n\left(x\right)\mathrm{if}a=b=0& \end{array}$

Short answer

Answer

The particular solution of the equation is given by $\left(6.24\right)$with $\mathbf{c}\mathbf{=}\mathbf{0}$.

Step-by-step solution

Given data.

he differential equation is $\left(\mathrm{D}-\mathrm{a}\right)\left(\mathrm{D}-\mathrm{b}\right)\mathrm{y}={\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right)$

${\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right)$is polynomial of degree

$\mathrm{c}=0$

$y_p$is role="math" localid="1653994318403" $\{\begin{array}{ll}a\mathrm{polynomial}{Q}_n\left(x\right)\mathrm{of}\mathrm{degree}n\mathrm{if}a\mathrm{and}& b\mathrm{are}\mathrm{both}\mathrm{different}\mathrm{from}\mathrm{zero};\\ {\mathrm{xQ}}_n\left(x\right)\mathrm{if}a\not\equiv 0,\mathrm{but}b=0,& \\ x^2{Q}_n\left(x\right)\mathrm{if}a=b=0& \end{array}$

Solution of differential equation.

$Q_n\left(x\right)=\sum _{}^{}{a}_n{x}^n$is a solution of the differential equation for a given $P_n=\sum _{}^{}{b}_n{x}^n$, if coefficients $a_n$can be found, so that $\left(D-a\right)\left(D-b\right)Q_n\left(x\right)=P_n\left(x\right)$.

Prove the particular solution of given differential equation.

For $a\ne b$

$$\begin{gathered} \left(D-a\right)\left(D-b\right){\mathbf{Q}}_{\mathbf{n}}\mathbf{=}{\mathbf{P}}_{\mathbf{n}} \\ \left(D^2-\left(a+b\right)D+\mathrm{ab}\right){\mathbf{Q}}_{\mathbf{n}}\mathbf{=}{\mathbf{P}}_{\mathbf{n}} \end{gathered}$$

But

$$\begin{gathered} Q_n\left(x\right)=\sum _{}^{}{a}_n{x}^n \\ \frac{d}{dx}{Q}_n=\sum _{}^{}n{a}_n{x}^{n-1}\frac{d^2}{d{x}^2}{Q}_n=\sum _{}^{}n\left(n-1\right)a_n{x}^{n-2} \end{gathered}$$

$\sum _{}^{}n\left(n-1\right)a_n{x}^{n-2}-\left(a+b\right)\sum _{}^{}n{a}_n{x}^{n-1}+ab\sum _{}^{}{a}_n{x}^n=\sum _{}^{}{b}_n{x}^n$

For $a\ne 0,b=0$

$$\begin{gathered} y_P=x{Q}_n\left(x\right) \\ {y}_P=Q_n+x{Q}_n=\sum _{}^{}\left(n+1\right)a_n{x}^n \\ {y}_P^{"}=2Q_{nn}^{"}\sum _{}^{}n\left(n+1\right)a_n{x}^{n-1} \end{gathered}$$

On putting

$\sum _{}^{}n\left(n+1\right)a_n{x}^{n-1}-\left(a+b\right)\sum _{}^{}\left(n+1\right)a_n{x}^n+abx\sum _{}^{}{a}_n{x}^n=\sum _{}^{}{b}_n{x}^{n}a=b=0D^2{y}_P=P_n{y}_P=x^2{Q}_n=x^2\sum _{}^{}{a}_n{x}^n$

Solve the equation further

$$\begin{gathered} y_P=2x{Q}_n \\ {y}_P^{"}=2Q_n^{\text{'}}+x^2{Q}_n^{"} \end{gathered}$$

$2\sum _{}^{}{a}_n{x}^n+4x\sum _{}^{}n{a}_n{x}^{n-1}+x^2\sum _{}^{}n\left(n+1\right)a_n{x}^{n-2}=\sum _{}^{}{b}_n{x}^n\sum _{}^{}{a}_n\left(2+3n+n^2\right)x^n=\sum _{}^{}{b}_n{x}^n$

Thus, proved