Opening the book…
- Solve Initial Value Problems using Laplace Transforms.
Laplace Transforms of Derivatives
In the rest of this chapter we’ll use the Laplace transform to solve initial value problems for constant coefficient first and 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.
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 8.1.6 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 in on the right side of Equation \ref{eq:8.3.2} converges as \(T\to\infty\) if \(s> s_0\). Therefore
\[\begin{aligned} \int^\infty_0 e^{-st}f'(t)\,dt&=-f(0)+s\int^\infty_0 e^{-st}f(t)\,dt\\[4pt] &=-f(0)+s{\cal L}(f),\end{aligned}\nonumber \]
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
\[\begin{aligned} \int^{t_i}_{t_{i-1}}e^{-st}f'(t)\,dt &= e^{-st}f(t)\Big|^{t_i}_{t_{i-1}}+s\int^{t_i}_{t_{i-1}}e^{-st}f(t)\,dt\\[4pt] &= e^{-st_i} f(t_i)- e^{-st_{i-1}}f(t_{i-1})+s\int^{t_i}_{t_{i-1}}e^{-st}f(t)\,dt.\end{aligned}\nonumber \]
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.
In Example 7.1.4 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
\[{\cal L}(-\omega\sin\omega t)=s {s\over s^2+\omega^2}-1=- {\omega^2\over s^2+\omega^2}.\nonumber \]
Therefore
\[{\cal L}(\sin\omega t)={\omega\over s^2+\omega^2},\nonumber \]
which agrees with the corresponding result obtained in 8.1.4.
In Section 2.4 we showed 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 7.3.1 , 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.
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
\[\label{eq:8.3.5} {\cal L}(f'')=s^2{\cal L}(f)-f'(0)-sf(0). \]
- Proof
-
Theorem 7.3.1 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 7.3.1 to \(g=f'\). Since \(g\) satisfies the hypotheses of Theorem 7.3.1 , 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
\[{\cal L}(f'')=s{\cal L}(f')-f'(0).\nonumber \]
Substituting Equation \ref{eq:8.3.4} into this yields Equation \ref{eq:8.3.5}.
Solving First Order Equations with the Laplace Transform
We’ll now use the Laplace transform to solve initial value problems for first order equations.
Use the Laplace transform to solve the initial value problem
\[\label{eq:7.3.6} y'+9y=t,\quad y(0)=1.\]
Solution
Taking Laplace transforms of both sides of the differential equation in Equation \ref{eq:7.3.6} yields
\[{\cal L}(y'+9y)={\cal L}(t)={1\over s^2},\nonumber \]
which we rewrite as
\[\label{eq:7.3.7} {\cal L}(y')+9{\cal L}(y)={1\over s^2}. \]
Now denote \({\cal L}(y)=Y(s)\). Theorem 7.3.2 and the initial conditions in Equation \ref{eq:7.3.6} imply that
\[{\cal L}(y')=sY(s)-y(0)=sY(s)-1\nonumber \]
Substituting into Equation \ref{eq:7.3.7} yields
\[\left(sY(s)-1\right)+9Y(s)={1\over s^2}.\nonumber \]
Therefore
\[\label{eq:7.3.8} (s+9)Y(s)={1\over s^2}+1, \]
so
\[Y(s)={1\over s^2(s+9)}+{1\over(s+9)}={(1+s^2)\over s^2(s+9)}.\nonumber \]
Using partial fraction decomposition, we get the partial fraction expansion
\[Y(s)={1\over9}{1\over s^2}-{1\over81}{1\over s}+{82\over81}{1\over s+9},\nonumber \]
and taking the inverse transform we get
\[{\cal L}^{-1}(Y(s))={1\over9}{\cal L}^{-1}\Big({1\over s^2}\Big)-{1\over81}{\cal L}^{-1}\Big({1\over s}\Big)+{82\over81}{\cal L}^{-1}\Big({1\over s+9}\Big),\nonumber \]
Therefore
\[y(t)={t\over9}e^{5t}-{1\over81}+{82\over81}e^{-9t} \nonumber \]
as the solution of Equation \ref{eq:7.3.6} on the interval \((0, \infty)\).
Solving Second Order Equations with the Laplace Transform
We’ll now use the Laplace transform to solve initial value problems for second order equations.
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 7.3.2 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
\[{\cal L}(y'')=s^2Y(s)-y'(0)-sy(0)=s^2Y(s)-3-2s.\nonumber \]
Substituting from the last two equations into Equation \ref{eq:8.3.7} yields
\[\left(s^2Y(s)-3-2s\right)-6\left(sY(s)-2\right)+5Y(s)={3\over s-2}.\nonumber \]
Therefore
\[\label{eq:8.3.8} (s^2-6s+5)Y(s)={3\over s-2}+(3+2s)+6(-2), \]
so
\[(s-5)(s-1)Y(s)={3+(s-2)(2s-9)\over s-2},\nonumber \]
and
\[Y(s)={3+(s-2)(2s-9)\over(s-2)(s-5)(s-1)}={2s^2-13s +21\over(s-2)(s-5)(s-1)}.\nonumber \]
Using partial fraction decomposition, we get the partial fraction expansion
\[Y(s)=-{1\over s-2}+{1\over2}{1\over s-5}+{5\over2}{1\over s-1},\nonumber \]
and taking the inverse transform we get
\[{\cal L}^{-1}(Y(s))=-{\cal L}^{-1}\Big({1\over s-2}\Big)+{1\over2}{\cal L}^{-1}\Big({1\over s-5}\Big)+{5\over2}{\cal L}^{-1}\Big({1\over s-1}\Big),\nonumber \]
Therefore
\[y(t)=-e^{2t}+{1\over2}e^{5t}+{5\over2}e^t \nonumber \]
as the solution of Equation \ref{eq:8.3.6} on the interval \((0, \infty)\).
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
\[{\cal L}( 2y''+3y'+y)={\cal L}\left(8e^{-2t}\right)={8\over s+2},\nonumber \]
which we rewrite as
\[\label{eq:8.3.15} 2{\cal L}(y'')+3{\cal L}(y')+{\cal L}(y)={8\over s+2}. \]
Now denote \({\cal L}(y)=Y(s)\). Theorem 7.3.2 and the initial conditions in Equation \ref{eq:8.3.14} imply that
\[{\cal L}(y')=sY(s)-y(0)=sY(s)+4\nonumber \]
and
\[{\cal L}(y'')=s^2Y(s)-y'(0)-sy(0)=s^2Y(s)-2+4s.\nonumber \]
Substituting from the last two equations into Equation \ref{eq:8.3.15} yields
\[\left(2s^2Y(s)-2+4s\right)+3\left(sY(s)+4\right)+Y(s)={8\over s+2}.\nonumber \]
Therefore
\[(2s+1)(s+1)Y(s)={8\over s+2}+2(2-4s)+3(-4).\nonumber \]
Solving for \(Y(s)\) yields
\[Y(s)={4\left(1-(s+2)(s+1)\right)\over (s+1/2)(s+1)(s+2)}=-{4s^2+12s+4\over (s+1/2)(s+1)(s+2)}.\nonumber \]
Partial fraction decompositions yields the partial fraction expansion
\[Y(s)={4\over3}{1\over s+1/2}-{8\over s+1}+{8\over3}{1\over s+2},\nonumber \]
and taking the inverse transform we get
\[{\cal L}^{-1}(Y(s))={4\over3}{\cal L}^{-1}\Big({1\over s+1/2}\Big)-8{\cal L}^{-1}\Big({1\over s+1}\Big)+{8\over3}{\cal L}^{-1}\Big({1\over s+2}\Big),\nonumber \]
Therefore the solution of Equation \ref{eq:8.3.14} over the interval \((0, \infty)\) is
\[y(t)={4\over3}e^{-t/2}-8e^{-t}+{8\over3}e^{-2t}\nonumber \]
(Figure 7.3.1 ).
This page titled 7.3: Solution of Initial Value Problems was last modified on Sun, 04 May 2025 21:30:50 GMT and is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform.
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