46.6.16 problem 66

Internal problem ID [9637]
Book : DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section : CHAPTER 7 THE LAPLACE TRANSFORM. 7.3.1 TRANSLATION ON THE s-AXIS. Page 297
Problem number : 66
Date solved : Tuesday, September 30, 2025 at 06:21:41 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

\begin{align*} y^{\prime \prime }+4 y&=\left \{\begin {array}{cc} 1 & 0\le t <1 \\ 0 & 1\le t \end {array}\right . \end{align*}

Using Laplace method With initial conditions

\begin{align*} y \left (0\right )&=0 \\ y^{\prime }\left (0\right )&=-1 \\ \end{align*}

✓ Maple. Time used: 0.318 (sec). Leaf size: 33

ode:=diff(diff(y(t),t),t)+4*y(t) = piecewise(0 <= t and t < 1,1,1 <= t,0); 
ic:=[y(0) = 0, D(y)(0) = -1]; 
dsolve([ode,op(ic)],y(t),method='laplace');
 

\[ y = \frac {\left (\left \{\begin {array}{cc} 1 & t <1 \\ \cos \left (2 t -2\right ) & 1\le t \end {array}\right .\right )}{4}-\frac {\sin \left (2 t \right )}{2}-\frac {\cos \left (2 t \right )}{4} \]

✓ Mathematica. Time used: 0.024 (sec). Leaf size: 65

ode=D[y[t],{t,2}]+4*y[t]==Piecewise[{{1,0<=t<1},{0,t>=1}}]; 
ic={y[0]==0,Derivative[1][y][0] ==-1}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

\begin{align*} y(t)&\to \begin {array}{cc} \{ & \begin {array}{cc} -\cos (t) \sin (t) & t\leq 0 \\ \frac {1}{4} (-\cos (2 t)-2 \sin (2 t)+1) & 0<t\leq 1 \\ \frac {1}{4} (\cos (2-2 t)-\cos (2 t)-2 \sin (2 t)) & \text {True} \\ \end {array} \\ \end {array} \end{align*}

✓ Sympy. Time used: 0.195 (sec). Leaf size: 22

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(-Piecewise((1, (t >= 0) & (t < 1)), (0, t >= 1)) + 4*y(t) + Derivative(y(t), (t, 2)),0) 
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): -1} 
dsolve(ode,func=y(t),ics=ics)
 

\[ y{\left (t \right )} = \begin {cases} \frac {1}{4} & \text {for}\: t \geq 0 \wedge t < 1 \\0 & \text {for}\: t \geq 1 \\\text {NaN} & \text {otherwise} \end {cases} - \frac {\sin {\left (2 t \right )}}{2} - \frac {\cos {\left (2 t \right )}}{4} \]