Application to Differential Equations

Consider the following example.

EXAMPLE 10.5.1   Solve the following Initial Value Problem:

$$a f^{\prime\prime}(t) + b f^\prime(t) +c f(t)= g(t) \; {\mbox{ with }}\; f(0) = f_0, \; f^\prime(0) = f_1.$$

Solution: Let ${\mathcal L}(g(t)) = G(s).$ Then

$$G(s) = a(s^2 F(s) - s f(0) - f^\prime(0)) + b ( s F(s) - f(0) ) +c F(s)$$

and the initial conditions imply

$$G(s) = (a s^2 + b s +c) F(s) - (a s + b) f_0 - a f_1.$$

Hence,

$$F(s) = \underbrace{\frac{G(s)}{a s^2 + b s + c}}_{ non-homogeneo... ... a s^2 + b s + c} + \frac{a f_1}{a s^2 + b s + c}}_{initial \;\; conditions}.$$ (10.5.1)

Now, if we know that $G(s)$ is a rational function of $s$ then we can compute $f(t)$ from $F(s)$ by using the method of PARTIAL FRACTIONS (see Subsection 10.3.1).

EXAMPLE 10.5.2  

1. Solve the IVP
   $$y^{\prime\prime} - 4 y^\prime - 5 y = f(t) = \left\{\begin{array... ...} \;\; 0 \leq t < 5 \\ t+5 & {\mbox{ if }} \;\; t \geq 5 \end{array}. \right.$$
   with $y(0) = 1$ and $y^\prime(0) = 4.$
   Solution: Note that $f(t) = t + U_5(t)$ . Thus,
   $${\mathcal L}(f(t)) = \frac{1}{s^2} + \frac{e^{-5s}}{s}.$$
   Taking Laplace transform of the above equation, we get
   $$\bigl(s^2 Y(s) - s y(0) - y^\prime(0)\bigr) - 4 \left(s Y(s) - y(0)\right) - 5 Y(s) = {\mathcal L}(f(t)) = \frac{1}{s^2} + \frac{e^{-5s}}{s}.$$
   Which gives
   

   $$Y(s) = \frac{s}{(s+1)(s-5)} + \frac{e^{-5s}}{s(s+1)(s-5)} + \frac{1}{s^2(s+1)(s-5)}$$

   $$= \frac{1}{6} \left[ \frac{5}{s-5} + \frac{1}{s+1} \right] + \frac{e^{-5s}}{30} \left[ -\frac{6}{s} + \frac{5}{s+1} + \frac{1}{s-5} \right]$$

   $$+ \frac{1}{150} \left[ -\frac{30}{s^2} + \frac{24}{s} - \frac{25}{s+1} + \frac{1}{s-5} \right].$$

   
   Hence,
   

   $$y(t) = \frac{5 e^{5t}}{6} + \frac{e^{-t}}{6} + U_5(t) \left[ -\frac{1}{5} + \frac{e^{-(t-5)}}{6} + \frac{e^{5(t-5)}}{30} \right]$$

   $$+ \frac{1}{150} \left[ - 30 t + 24 - 25 e^{-t} + e^{5t} \right].$$

   
   

Remark 10.5.3   Even though $f(t)$ is a DISCONTINUOUS function at $t = 5,$ the solution $y(t)$ and $y^\prime(t)$ are continuous functions of $t$ , as $y^{\prime\prime}$ exists. In general, the following is always true:
Let $y(t)$ be a solution of $a y^{\prime\prime} + b y^\prime + c y = f(t).$ Then both $y(t)$ and $y^\prime(t)$ are continuous functions of time.

EXAMPLE 10.5.4  

1. Consider the IVP $t y^{\prime\prime}(t) + y^\prime(t) + t y(t) = 0,$ with $y(0) = 1$ and $y^\prime(0) = 0.$ Find ${\mathcal L}(y(t)).$
   Solution: Applying Laplace transform, we have
   
   $$- \frac{d}{ds}\left[ s^2 Y(s) - s y(0) - y^\prime (0) \right] + (s Y(s) - y(0)) - \frac{d}{ds} Y(s) = 0.$$
   Using initial conditions, the above equation reduces to
   $$\frac{d}{ds} \left[ (s^2 +1) Y(s) - s \right] - s Y(s) + 1 = 0.$$
   This equation after simplification can be rewritten as
   $$\frac{Y^\prime(s)}{Y(s)} = - \frac{s}{s^2 + 1}.$$
   Therefore, $$Y(s) = a (1+s^2)^{-\frac{1}{2}}.$$ From Example 10.4.2.1, we see that $a = 1$ and hence
   $$\displaystyle Y(s) = (1+s^2)^{-\frac{1}{2}}.$$

2. Show that $y(t) = \displaystyle \int_0^t f(\tau) g(t -\tau) d \tau$ is a solution of
   $$y^{\prime\prime}(t) + a y^\prime(t) + b y(t) = f(t), \;\;\; {\mbox{ with }} \;\; y(0) = y^\prime(0) = 0;$$
   where ${\mathcal L}[g(t)] = \displaystyle\frac{1}{s^2 + a s + b}.$
   Solution: Here, $Y(s) = \displaystyle\frac{F(s)}{s^2 + a s + b} = F(s)\cdot \frac{1}{s^2 + as + b}.$ Hence,
   $$y(t) = \displaystyle (f \star g)(t) = \displaystyle\int_0^t f(\tau) g(t -\tau) d \tau.$$

3. Show that $y(t) = \displaystyle\frac{1}{a} \int_0^t f(\tau) \sin (a(t -\tau)) d \tau$ is a solution of
   $$y^{\prime\prime}(t) + a^2 y(t) = f(t), \;\;\; {\mbox{ with }} \;\; y(0) = y^\prime(0) = 0.$$
   
   Solution: Here, $Y(s) = \displaystyle\frac{F(s)}{s^2 + a^2} = \frac{1}{a} \left( F(s)\cdot \frac{a}{s^2 + a^2}\right).$ Hence,
   $$y(t) = \displaystyle\frac{1}{a} f(t) \star \sin (at) = \displaystyle\frac{1}{a} \int_0^t f(\tau) \sin (a(t -\tau)) d \tau.$$

4. Solve the following IVP.
   $$y^\prime(t) = \int_0^t y(\tau) d \tau + t - 4 \sin t, \;\; {\mbox{ with }} \;\; y(0) = 1.$$
   
   Solution: Taking Laplace transform of both sides and using Theorem 10.3.5, we get
   $$s Y(s) - 1 = \frac{Y(s)}{s} + \frac{1}{s^2} - 4 \frac{1}{s^2 + 1}.$$
   Solving for $Y(s),$ we get
   $$Y(s) = \frac{s^2 -1}{s(s^2 + 1)} = \frac{1}{s} - 2\frac{1}{s^2 + 1}.$$
   So,
   $$y(t) = 1 - 2 \int_0^t \sin (\tau) d \tau = 1 + 2 (\cos t -1) = 2 \cos t - 1.$$