9.3: Solutions of Initial Value Problems

Laplace Transforms of Derivatives

In the rest of this chapter we’ll use the Laplace transform to solve initial value problems for constant coefficient second order equations. To do this, we must know how the Laplace transform of \(f'\) is related to the Laplace transform of \(f\). The next theorem answers this question.

Theorem 9.3.1

Suppose \(f\) is continuous on \([0,\infty)\) and of exponential order \(s_0\), and \(f'\) is piecewise continuous on \([0,\infty).\) Then \(f\) and \(f'\) have Laplace transforms for \(s > s_0,\) and

\[\label{eq:8.3.1} {\cal L}(f')=s {\cal L}(f)-f(0).\]

Proof

We know from Theorem 9.1.2 that \({\cal L}(f)\) is defined for \(s>s_0\). We first consider the case where \(f'\) is continuous on \([0,\infty)\). Integration by parts yields

\[ \begin{align} \int^T_0 e^{-st}f'(t)\,dt &= e^{-st}f(t)\Big|^T_0+s \int^T_0e^{-st}f(t)\,dt \nonumber \\[4pt] &= e^{-sT}f(T)-f(0)+s\int^T_0 e^{-st}f(t)\,dt \label{eq:8.3.2} \end{align} \]

for any \(T>0\). Since \(f\) is of exponential order \(s_0\), \(\displaystyle \lim_{T\to \infty}e^{-sT}f(T)=0\) and the integral on the right side of Equation \ref{eq:8.3.2} converges as \(T\to\infty\) if \(s> s_0\). Therefore

\[\begin{aligned} \int^\infty_0 e^{-st}f'(t)\,dt&=-f(0)+s\int^\infty_0 e^{-st}f(t)\,dt\\ &=-f(0)+s{\cal L}(f),\end{aligned}\nonumber \]

which proves Equation \ref{eq:8.3.1}.

Suppose \(T>0\) and \(f'\) is only piecewise continuous on \([0,T]\), with discontinuities at \(t_1 < t_2 <\cdots < t_{n-1}\). For convenience, let \(t_0=0\) and \(t_n=T\). Integrating by parts yields

\[\begin{aligned} \int^{t_i}_{t_{i-1}}e^{-st}f'(t)\,dt &= e^{-st}f(t)\Big|^{t_i}_{t_{i-1}}+s\int^{t_i}_{t_{i-1}}e^{-st}f(t)\,dt\\ &= e^{-st_i} f(t_i)- e^{-st_{i-1}}f(t_{i-1})+s\int^{t_i}_{t_{i-1}}e^{-st}f(t)\,dt.\end{aligned}\nonumber\]

Summing both sides of this equation from \(i=1\) to \(n\) and noting that

\[\left(e^{-st_1}f(t_1)-e^{-st_0}f(t_0)\right)+\left(e^{-st_2} f(t_2)-e^{-st_1}f(t_1)\right) +\cdots+\left(e^{-st_N}f(t_N)-e^{-st_{N-1}}f(t_{N-1})\right) \nonumber\]

\[=e^{-st_N}f(t_N)-e^{-st_0}f(t_0)=e^{-sT}f(T)-f(0) \nonumber\]

yields Equation \ref{eq:8.3.2}, so Equation \ref{eq:8.3.1} follows as before.

Example 9.3.1

In section 9.1 we saw that

\[{\cal L}(\cos\omega t)={s\over s^2+\omega^2}. \nonumber\]

Solution

Applying Equation \ref{eq:8.3.1} with \(f(t)=\cos\omega t\) shows that

\[{\cal L}\left(\sin(\omega t)\right)={\cal L}\left({-1\over \omega}{d\over dt}\cos(\omega t)\right)={-1\over \omega}{\cal L}\left({d\over dt}\cos(\omega t)\right)={-1\over \omega}\left(s{\cal L}(\cos(\omega t)-\cos(0)\right)={-1\over \omega}\left(s {s\over s^2+\omega^2}-1\right)={-1\over \omega} {-\omega^2\over s^2+\omega^2}={\omega\over s^2+\omega^2}.\nonumber\]

Therefore

\[{\cal L}(\sin\omega t)={\omega\over s^2+\omega^2},\nonumber\]

which agrees with the corresponding result obtained in section 9.1.

Using separation of variables from Chapter 2 we can show that the solution of the initial value problem

\[\label{eq:8.3.3} y'=ay, \quad y(0)=y_0,\]

is \(y=y_0e^{at}\). We’ll now obtain this result by using the Laplace transform.

Let \(Y(s)={\cal L}(y)\) be the Laplace transform of the unknown solution of Equation \ref{eq:8.3.3}. Taking Laplace transforms of both sides of Equation \ref{eq:8.3.3} yields

\[{\cal L}(y')={\cal L}(ay),\nonumber\]

which, by Theorem 9.3.1 , can be rewritten as

\[s{\cal L}(y)-y(0)=a{\cal L}(y),\nonumber\]

or

\[sY(s)-y_0=aY(s).\nonumber\]

Solving for \(Y(s)\) yields

\[Y(s)={y_0\over s-a},\nonumber\]

so

\[y={\cal L}^{-1}(Y(s))={\cal L}^{-1}\left({y_0\over s-a}\right)=y_0{\cal L}^{-1}\left({1\over s-a}\right)=y_0e^{at},\nonumber\]

which agrees with the known result.

We need the next theorem to solve second order differential equations using the Laplace transform.

Theorem 9.3.2

Suppose \(f\) and \(f'\) are continuous on \([0,\infty)\) and of exponential order \(s_0,\) and that \(f''\) is piecewise continuous on \([0,\infty).\) Then \(f\), \(f'\), and \(f''\) have Laplace transforms for \(s > s_0\),

\[\label{eq:8.3.4} {\cal L}(f')=s {\cal L}(f)-f(0),\]

and

\[\label{eq:8.3.5} {\cal L}(f'')=s^2{\cal L}(f)-sf(0)-f'(0).\]

Proof

Theorem 9.3.1 implies that \({\cal L}(f')\) exists and satisfies Equation \ref{eq:8.3.4} for \(s>s_0\). To prove that \({\cal L}(f'')\) exists and satisfies Equation \ref{eq:8.3.5} for \(s>s_0\), we first apply Theorem 9.3.1 to \(g=f'\). Since \(g\) satisfies the hypotheses of Theorem 9.3.1 , we conclude that \({\cal L}(g')\) is defined and satisfies

\[{\cal L}(g')=s{\cal L}(g)-g(0)\nonumber\]

for \(s>s_0\). However, since \(g'=f''\), this can be rewritten as

\[{\cal L}(f'')=s{\cal L}(f')-f'(0)=s(s{\cal L}(f)-f(0))-f'(0)=s^2{\cal L}(f)-sf(0)-f'(0),\nonumber\]

which is Equation \ref{eq:8.3.5}.

Solving Second Order Equations with the Laplace Transform

We’ll now use the Laplace transform to solve initial value problems for second order equations.

Example 9.3.2

Use the Laplace transform to solve the initial value problem

\[\label{eq:8.3.6} y''-6y'+5y=3e^{2t},\quad y(0)=2, \quad y'(0)=3.\]

Solution

Taking Laplace transforms of both sides of the differential equation in Equation \ref{eq:8.3.6} yields

\[{\cal L}(y''-6y'+5y)={\cal L}\left(3e^{2t}\right)={3\over s-2},\nonumber\]

which we rewrite as

\[\label{eq:8.3.7} {\cal L}(y'')-6{\cal L}(y')+5{\cal L}(y)={3\over s-2}.\]

Now denote \({\cal L}(y)=Y(s)\). Theorem 9.3.2 and the initial conditions in Equation \ref{eq:8.3.6} imply that

\[{\cal L}(y')=sY(s)-y(0)=sY(s)-2\nonumber\]

and

\[{\cal L}(y'')=s^2Y(s)-sy(0)-y'(0)=s^2Y(s)-2s-3.\nonumber\]

Substituting from the last two equations into Equation \ref{eq:8.3.7} yields

\[\left(s^2Y(s)-2s-3\right)-6\left(sY(s)-2\right)+5Y(s)={3\over s-2}.\nonumber\]

Therefore

\[\label{eq:8.3.8} (s^2-6s+5)Y(s)={3\over s-2}+(2s+3)+6(-2),\]

so

\[(s-5)(s-1)Y(s)={3+(s-2)(2s-9)\over s-2},\nonumber\]

and

\[Y(s)={3+(s-2)(2s-9)\over(s-2)(s-5)(s-1)}.\nonumber\]

Partial fraction expansion gives us

\[Y(s)=-{1\over s-2}+{1\over2}{1\over s-5}+{5\over2}{1\over s-1},\nonumber\]

and taking the inverse transform of this yields

\[y=-e^{2t}+{1\over2}e^{5t}+{5\over2}e^t \nonumber\]

as the solution of Equation \ref{eq:8.3.6}.

From here on out we will solve these equations in a more streamlined manner, as the next example illustrates. 

Example 9.3.3

Use the Laplace transform to solve the initial value problem

\[\label{eq:8.3.14} 2y''+3y'+y=8e^{-2t}, \quad y(0)=-4,\; y'(0)=2.\]

Solution

Taking Laplace transforms of both sides of the differential equation in Equation \ref{eq:8.3.14} yields

\[2s^2Y(s)-2sy(0)-2y'(0)+3sY(s)-3y(0)+Y(s)={8\over s+2}.\nonumber\]

Moving all \(Y(s)\) terms to the left and all other terms to the right yields

\[(2s^2+3s+1)Y(s)={8\over s+2}+2(2-4s)+3(-4).\nonumber\]

which becomes

\[(2s+1)(s+1)Y(s)={8\over s+2}+2(2-4s)+3(-4).\nonumber\]

Solving for \(Y(s)\) yields

\[Y(s)={4\left(1-(s+2)(s+1)\right)\over (s+1/2)(s+1)(s+2)}.\nonumber\]

The partial fraction expansion gives us

\[Y(s)={4\over3}{1\over s+1/2}-{8\over s+1}+{8\over3}{1\over s+2},\nonumber\]

so the solution of Equation \ref{eq:8.3.14} is

\[y={\cal L}^{-1}(Y(s))={4\over3}e^{-t/2}-8e^{-t}+{8\over3}e^{-2t}\nonumber\]

(Figure 9.3.1 ).

Figure 9.3.1 : \(y=\frac{4}{3}e^{-t/2}-8e^{-t}+\frac{8}{3}e^{-2t}\)