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7.2 Inverse Transforms and Transforms of Derivatives: Class Notes

7.2a

Theorem 7.2.1: Some Inverse Transforms

• (a) ${\displaystyle 1={ℒ}^{-1}\left\{\frac{1}{s}\right\}}$
• (b) ${\displaystyle t^n={ℒ}^{-1}\left\{\frac{n!}{s^{n+1}}\right\}}$, $n=1,2,3,\dots$
• (c) ${\displaystyle e^{at}={ℒ}^{-1}\left\{\frac{1}{s-a}\right\}}$
• (d) ${\displaystyle \sin kt={ℒ}^{-1}\left\{\frac{k}{s^2+k^2}\right\}}$
• (e) ${\displaystyle \cos kt={ℒ}^{-1}\left\{\frac{s}{s^2+k^2}\right\}}$
• (f) ${\displaystyle \sinh kt={ℒ}^{-1}\left\{\frac{k}{s^2-k^2}\right\}}$
• (g) ${\displaystyle \cosh kt={ℒ}^{-1}\left\{\frac{s}{s^2-k^2}\right\}}$

Example

Find the inverse Laplace transform.

${\displaystyle {ℒ}^{-1}\left\{\frac{1}{s^4}\right\}}$

Solution

This matches part (b) with $n=3\Rightarrow n!=6$. The numerator needs to be $6$, so multiply and divide by 6:

$${\displaystyle {ℒ}^{-1}\left\{\frac{1}{s^4}\right\}=\frac{1}{6}{ℒ}^{-1}\left\{\frac{6}{s^4}\right\}=\frac{1}{6}{t}^3}$$

${\displaystyle {ℒ}^{-1}\left\{\frac{10s}{s^2+16}\right\}}$

Solution

The $s^2+16$ in the denominator means a trig function, with $k=4$. The $s$ in the numerator makes it cosine:

$${\displaystyle {ℒ}^{-1}\left\{\frac{10s}{s^2+16}\right\}=10{ℒ}^{-1}\left\{\frac{s}{s^2+16}\right\}=10\cos 4t}$$

${\displaystyle {ℒ}^{-1}\left\{\frac{s+1}{s^2+2}\right\}}$

Solution

Here $k=\sqrt{2}$. Split the fraction; the second piece needs $k=\sqrt{2}$ in the numerator, so multiply and divide by $\sqrt{2}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {ℒ}^{-1}\left\{\frac{s+1}{s^2+2}\right\}}& {\displaystyle ={ℒ}^{-1}\{\frac{s}{s^2+2}+\frac{1}{\sqrt{2}}·\frac{1·\sqrt{2}}{s^2+2}\}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\cos \sqrt{2}t+\frac{1}{\sqrt{2}}\sin \sqrt{2}t}\hfill \end{array}}$$

${\displaystyle {ℒ}^{-1}\left\{\frac{s+1}{s^2-4s}\right\}}$

Solution

The denominator factors as $s(s-4)$, so use partial fraction decomposition (PFD):

$${\displaystyle \frac{s+1}{s(s-4)}=\frac{A}{s}+\frac{B}{s-4}\Rightarrow s+1=A(s-4)+Bs}$$

$s=0$: $1=-4A$, so ${\displaystyle A=-\frac{1}{4}}$. $s=4$: $5=4B$, so ${\displaystyle B=\frac{5}{4}}$.

$${\displaystyle {ℒ}^{-1}\{\frac{-1/4}{s}+\frac{5/4}{s-4}\}=-\frac{1}{4}+\frac{5}{4}{e}^{4t}}$$

7.2b

Example

Use the Laplace transform to solve the Initial-Value Problem.

$${\displaystyle 2\frac{dy}{dt}+y=0,y(0)=-3}$$

Solution

Plan: Apply the Laplace transform to both sides.

$${\displaystyle 2ℒ\left\{\frac{dy}{dt}\right\}+ℒ\{y(t)\}=ℒ\left\{0\right\}}$$

The transform $ℒ$ takes $y(t)$ to $Y(s)$, and the inverse transform ${ℒ}^{-1}$ takes $Y(s)$ back to $y(t)$. To transform the equation, we need the transform of a derivative.

Transforms of derivatives

$${\displaystyle ℒ\{f\prime (t)\}={\int }_0^{\infty }{e}^{-st}f\prime (t)dt}$$

Integrate by parts with $u=e^{-st},dv=f\prime (t)dt$, so $du=-s{e}^{-st}dt,v=f(t)$.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{f\prime (t)\}}& {\displaystyle ={\frac{f(t)}{e^{st}}|}_0^{\infty }+s{\int }_0^{\infty }{e}^{-st}f(t)dt}\hfill \\ \hfill {\displaystyle }& {\displaystyle =-f(0)+sF(s)}\hfill \end{array}}$$

If $f(t)$ is of exponential order, ${\displaystyle \frac{f(t)}{e^{st}}\to 0}$ as $t\to \infty$ (for large enough $s$), so only the lower limit contributes. So

$$ℒ\{f\prime (t)\}=sF(s)-f(0)$$

The value $f(0)$ comes from the initial condition.

Plan (cont): Solve for $Y(s)$ and use ${ℒ}^{-1}$.

Back to the example: transform ${\displaystyle 2\frac{dy}{dt}+y=0}$ and substitute $y(0)=-3$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle 2\left(sY(s)-y(0)\right)+Y(s)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle 2sY(s)+6+Y(s)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (2s+1)Y(s)}& {\displaystyle =-6}\hfill \\ \hfill {\displaystyle Y(s)}& {\displaystyle =-\frac{6}{2s+1}=-\frac{3}{s+1/2}}\hfill \\ \hfill {\displaystyle y(t)}& {\displaystyle =-3e^{-t/2}}\hfill \end{array}}$$

The second derivative. The same steps give the transform of $f″$:

$${\displaystyle ℒ\{f″(t)\}={\int }_0^{\infty }{e}^{-st}f″(t)dt}$$

Integrate by parts with $u=e^{-st},dv=f″(t)dt$, so $du=-s{e}^{-st}dt,v=f\prime (t)$.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{f″(t)\}}& {\displaystyle ={\frac{f\prime (t)}{e^{st}}|}_0^{\infty }+s{\int }_0^{\infty }{e}^{-st}f\prime (t)dt}\hfill \\ \hfill {\displaystyle }& {\displaystyle =-f\prime (0)+sℒ\{f\prime (t)\}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =-f\prime (0)+s\left(sF(s)-f(0)\right)}\hfill \end{array}}$$

$$ℒ\{f″(t)\}=s^{2}F(s)-sf(0)-f\prime (0)$$

Guess: $ℒ\{f‴(t)\}=s^{3}F(s)-s^{2}f(0)-sf\prime (0)-f″(0)$

7.2c

Theorem: Transform of a Derivative

If $f,f\prime ,\dots ,f^{(n-1)}$ are continuous on $[0,\infty )$ and are of exponential order and if $f^{(n)}(t)$ is piecewise continuous on $[0,\infty )$, then

$$ℒ\{f^{(n)}(t)\}=s^{n}F(s)-s^{n-1}f(0)-s^{n-2}f\prime (0)-\cdots -f^{(n-1)}(0),$$

where $F(s)=ℒ\{f(t)\}$.

Example

Use the Laplace transform to solve the Initial-Value Problem.

$$y″-4y\prime =6e^{3t}-3e^{-t},y(0)=1,y\prime (0)=-1$$

Solution

Transform each term, matching colors to the equation:

$${\displaystyle s^{2}Y(s)-sy(0)-y\prime (0)-4(sY(s)-y(0))=\frac{6}{s-3}-\frac{3}{s+1}}$$

Substitute $y(0)=1$ and $y\prime (0)=-1$, then solve for $Y(s)$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle (s^2-4s)Y(s)-s+1+4}& {\displaystyle =\frac{6}{s-3}-\frac{3}{s+1}}\hfill \\ \hfill {\displaystyle (s^2-4s)Y(s)}& {\displaystyle =\frac{6}{s-3}-\frac{3}{s+1}+s-5}\hfill \\ \hfill {\displaystyle (s^2-4s)Y(s)}& {\displaystyle =\frac{6(s+1)-3(s-3)+(s-5)(s+1)(s-3)}{(s+1)(s-3)}}\hfill \\ \hfill {\displaystyle (s^2-4s)Y(s)}& {\displaystyle =\frac{6s+6-3s+9+s^3-7s^2+7s+15}{(s+1)(s-3)}}\hfill \\ \hfill {\displaystyle Y(s)}& {\displaystyle =\frac{s^3-7s^2+10s+30}{s(s-4)(s+1)(s-3)}}\hfill \end{array}}$$

Partial fractions:

$${\displaystyle \frac{s^3-7s^2+10s+30}{s(s-4)(s+1)(s-3)}=\frac{A}{s}+\frac{B}{s-4}+\frac{C}{s+1}+\frac{D}{s-3}}$$

$$\begin{array}{cc}\hfill {\displaystyle s^3-7s^2+10s+30}& {\displaystyle =A(s-4)(s+1)(s-3)+Bs(s+1)(s-3)}\hfill \\ \hfill {\displaystyle }& {\displaystyle +Cs(s-4)(s-3)+Ds(s-4)(s+1)}\hfill \end{array}$$

Substitute each root of the denominator; every term but one becomes zero:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle s=0:}& {\displaystyle 30=A(-4)(1)(-3)\Rightarrow A=\frac{5}{2}}\hfill \\ \hfill {\displaystyle s=4:}& {\displaystyle 64-112+40+30=B(4)(5)(1)\Rightarrow B=\frac{11}{10}}\hfill \\ \hfill {\displaystyle s=-1:}& {\displaystyle -1-7-10+30=C(-1)(-5)(-4)\Rightarrow C=-\frac{3}{5}}\hfill \\ \hfill {\displaystyle s=3:}& {\displaystyle 27-63+30+30=D(3)(-1)(4)\Rightarrow D=-2}\hfill \end{array}}$$

$${\displaystyle Y(s)=\frac{5/2}{s}+\frac{11/10}{s-4}-\frac{3/5}{s+1}-\frac{2}{s-3}}$$