Question

(a) Laguerre’s differential equation $\mathit{t}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{-}\mathit{t}\mathbf{)}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{n}\mathit{y}\mathbf{=}\mathbf{0}$is known to possess polynomial solutions whenis a nonnegative integer. These solutions are naturally called Laguerre polynomials and are denoted by L_n(t). Find y= L_n(t)for n = 0,1,2,3,4if it is known that L_n(0) = 1.

(b) Show that $\mathcal{L}\left\{\frac{e^t}{n!}\frac{d^n}{d{t}^n}{t}^n{e}^{-t}\right\}\mathbf{=}\mathit{Y}\mathbf{\left(}\mathit{s}\mathbf{\right)}$where $\mathit{Y}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}\mathcal{L}\mathbf{\left\{}\mathit{y}\mathbf{\right\}}$and y = L_n(t)is a polynomial solution of a DE in part (a) . Conclude that

${\mathit{L}}_{\mathbf{n}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{t}}}{\mathbf{n}\mathbf{!}}\frac{{\mathbf{d}}^{\mathbf{n}}}{\mathbf{d}{\mathbf{t}}^{\mathbf{n}}}{\mathit{t}}^{\mathbf{n}}{\mathit{e}}^{\mathbf{-}\mathbf{t}}\mathbf{,}\mathbf{}\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{\dots }$

This last relation for generating the Laguerre polynomials is the analogue of Rodrigues’ formula for the Legendre polynomials. See (36) in Section 6.4.

Short answer

(a) The solution for y = L_n(t), when n = 0,1,2,3,4 is,

$n=0:L_0\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}=1$

$n=1:L_1\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{s-1}{s^2}\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{s}-\frac{1}{s^2}\right\}=1-t$

$n=2:L_2\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{{(s-1)}^2}{s^3}\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{s}-\frac{2}{s^2}+\frac{1}{s^3}\right\}=1-2t+\frac{1}{2}{t}^2$

$n=3:L_3\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{{(s-1)}^3}{s^4}\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{s}-\frac{3}{s^2}+\frac{3}{s^3}-\frac{1}{s^4}\right\}=1-3t+\frac{3}{2}{t}^2-\frac{1}{6}{t}^3$

$n=4:L_4\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{{(s-1)}^4}{s^5}\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{s}-\frac{4}{s^2}+\frac{6}{s^3}-\frac{4}{s^4}+\frac{1}{s^5}\right\}$$=1-4t+3t^2-\frac{2}{3}{t}^3+\frac{1}{24}{t}^4$

(b) The equation $L_n\left(t\right)=\frac{e^t}{n!}\frac{d^n}{d{t}^n}{t}^n{e}^{-t},n=0,1,2,\dots$can be concluded by substituting the function values.

Step-by-step solution

Define derivatives of transforms theorem:

If $F\left(s\right)=\mathcal{L}\left\{f\right(t\left)\right\}$and n = 1,2,3,..... then

$\mathcal{L}\left\{t^{n}f\left(t\right)\right\}=(-1{)}^n\frac{d^n}{d{s}^n}F\left(s\right)$

Apply the Laplace transform to find the required equation:

Hence, we know

$(*)\left\{\begin{array}{ll}L\left\{y\text{'}\text{'}\right\}=& s^{2}L\left\{y\right\}-sy\left(0\right)-y\text{'}\left(0\right)\\ L\left\{y\text{'}\right\}=& sL\left\{y\right\}-y\left(0\right)\end{array}\right.$

Use Laplace transform on both sides,

$\mathcal{L}\left\{0\right\}=\mathcal{L}\left\{ly\text{'}\text{'}\right\}+\mathcal{L}\left\{y\text{'}\right\}-\mathcal{L}\left\{ly\text{'}\right\}+\mathcal{L}\left\{ny\right\}$

Thus, apply the derivatives of transforms theorem,

$\text{=}-\frac{\text{d}}{\text{d}s}\left[\mathcal{L}\left\{y\text{'}\text{'}\right\}\right]+\mathcal{L}\left\{y\text{'}\right\}+\frac{\text{d}}{\text{d}s}\left[\mathcal{L}\left\{y\text{'}\right\}\right]+\mathcal{L}\left\{ny\right\}$

$=┴\left(\right(*\left)\right)-\frac{d}{ds}[s^{2}Ly-sy(0)-y^{\text{'}}(0\left)\right]+sLy-y\left(0\right)+\frac{d}{ds}[sLy-y(0\left)\right]+Lny$

$=-\frac{d}{ds}\left[s^{2}Ly\right]+sLy+\frac{d}{ds}\left[sLy\right]+nLy\hspace{1em}\left(y\right(0)=0,y^{\text{'}}(0)=0)$

Substitute,$\mathcal{L}\left\{y\right\}=Y\left(s\right)$in the equation

$0=-\frac{\text{d}}{\text{d}s}\left[s^{2}Y\left(s\right)\right]+sY\left(s\right)+\frac{\text{d}}{\text{d}s}\left[sY\right(s\left)\right]+nY\left(s\right)$

$=-2sY\left(s\right)-s^{2}Y\text{'}\left(s\right)+sY\left(s\right)+Y\left(s\right)+sY\text{'}\left(s\right)+nY\left(s\right)$

$=Y\text{'}\left(s\right)\left(-s^2+s\right)+Y\left(s\right)(-s+1+n)$

Rewrite the differential equation as,

$\frac{\text{d}Y\left(s\right)}{\text{d}s}\left(s^2-s\right)=(n+1-s)Y\left(s\right)$

$\frac{\text{d}Y\left(s\right)}{Y\left(s\right)}=\frac{n+1-s}{s^2-s}\text{d}s$

Hence, to solve $\frac{n+1-s}{s(s-1)}$, use the partial fraction,

$\star \frac{n+1-s}{s(s-1)}=\frac{A}{s}+\frac{B}{s-1}=\frac{A(s-1)+Bs}{s(s-1)}$-------(2)

Determine the values of A and B to find the solution:

Since, the fractions have same denominators, the numerators will be equal.

$n+1-s=As-A+Bs$

n + 1 - s = (A + B)s - A

Thus, compare the coefficients

$$\begin{gathered} A+B=-1 \\ -A=n+1 \end{gathered}$$

$$\begin{gathered} B=-1-A \\ A=-(n+1) \end{gathered}$$

B = n

A = -(n+1)

Substitute the values of A and B in equation

$\frac{n+1-s}{s(s-1)}=-\frac{n+1}{s}+\frac{n}{s-1}$

Rewrite the equation (1) as,

$\frac{\text{d}Y\left(s\right)}{Y\left(s\right)}=\left[\frac{n}{s-1}-\frac{n+1}{s}\right]\text{d}s$

Hence, apply integration on both sides of the equation

$\int \frac{\text{d}Y\left(s\right)}{Y\left(s\right)}=\int \left[\frac{n}{s-1}-\frac{n+1}{s}\right]\text{d}s$

$ln\left|Y\right(s\left)\right|=n\ln |s-1|-(n+1)\ln \left|s\right|+\ln \left|c\right|$

$\ln \left|Y\right(s\left)\right|=\ln \left|\frac{{(s-1)}^n}{s^{n+1}}\right|+\ln \left|c\right|$

$\ln \left|Y\right(s\left)\right|=\ln \left|\frac{{(s-1)}^n}{s^{n+1}}·c\right|$

$Y\left(s\right)=\frac{{(s-1)}^n}{s^{n+1}}·c$

Assume c = 1 ,

$Y\left(s\right)=\frac{{(s-1)}^n}{s^{n+1}}$

Find the solution for the values

Substitute $y\left(t\right)=L_n\left(t\right)$,and$\mathcal{L}\left\{y\right(t\left)\right\}=Y\left(s\right)$

$L_n\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{{(s-1)}^n}{s^{n+1}}\right\}$

Therefore, for n = 1,2,3,4.... the required values are,

$n=0:L_0\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}=1$

$n=1:L_1\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{s-1}{s^2}\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{s}-\frac{1}{s^2}\right\}=1-t$

$n=2:L_2\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{{(s-1)}^2}{s^3}\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{s}-\frac{2}{s^2}+\frac{1}{s^3}\right\}=1-2t+\frac{1}{2}{t}^2$

$n=3:L_3\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{{(s-1)}^3}{s^4}\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{s}-\frac{3}{s^2}+\frac{3}{s^3}-\frac{1}{s^4}\right\}=1-3t+\frac{3}{2}{t}^2-\frac{1}{6}{t}^3$

$n=4:L_4\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{{(s-1)}^4}{s^5}\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{s}-\frac{4}{s^2}+\frac{6}{s^3}-\frac{4}{s^4}+\frac{1}{s^5}\right\}=1-4t+3t^2-\frac{2}{3}{t}^3+\frac{1}{24}{t}^4$

Consider the function and evaluate the equation:

(b)

Hence the function, $f\left(t\right)=t^n{e}^{-t}$

Thus, localid="1668173052064" $f^{\left(k\right)}\left(0\right)=0\text{for}k=0,1,2,\dots ,n-1$and $f^{\left(n\right)}\left(0\right)=n!$.

$\frac{1}{n!}\mathcal{L}\left\{e^t\frac{{\text{d}}^n}{\text{d}{t}^n}{t}^n{e}^{-t}\right\}=\frac{1}{n!}\mathcal{L}\left\{e^t{f}^n\left(t\right)\right\}$

$={\left.\frac{1}{n!}\mathcal{L}\left\{f^{\left(n\right)}\left(t\right)\right\}\right|}_{s\to s-1}$

Expand the function,

$={\left.\frac{1}{n!}\left[s^n\mathcal{L}\left\{t^n{e}^{-t}\right\}-s^{n-1}f\left(0\right)-s^{n-2}f\text{'}\left(0\right)-\cdots -f^{(n-1)}\left(0\right)\right]\right|}_{n\to s-1}$

$={\left.\frac{1}{n!}\left[s^n\mathcal{L}\left\{l^n{e}^{-t}\right\}-0-0-\cdots -0\right]\right|}_{s>s\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 1}$

$={\left.\frac{1}{n!}\left[s^n\mathcal{L}\left\{t^n{e}^{-t}\right\}\right]\right|}_{s\to s-1}$

$={\left.\frac{1}{n!}\left[s^n\frac{n!}{{(s+1)}^{n+1}}\right]\right|}_{s\to s-1}$

Solve,

$=\frac{{(s-1)}^n}{s^{n+1}}$

= Y(s)

Since, $Y=\mathcal{L}\left\{L_n\left(t\right)\right\}$,

$L_n\left(t\right)=\frac{1}{n!}{e}^t\frac{{\text{d}}^n}{\text{d}{t}^n}{t}^n{e}^{-t},n=0,1,2,\dots$

Therefore, it is concluded that

$L_n\left(t\right)=\frac{1}{n!}{e}^t\frac{{\text{d}}^n}{\text{d}{t}^n}{t}^n{e}^{-t},n=0,1,2,\dots$