7.20.9 problem 38

Internal problem ID [563]
Book : Elementary Differential Equations. By C. Henry Edwards, David E. Penney and David Calvis. 6th edition. 2008
Section : Chapter 4. Laplace transform methods. Section 4.4 (Derivatives, Integrals and products of transforms). Problems at page 303
Problem number : 38
Date solved : Monday, January 27, 2025 at 02:54:51 AM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

\begin{align*} x^{\prime \prime }+4 x^{\prime }+13 x&=f \left (t \right ) \end{align*}

Using Laplace method With initial conditions

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

✓ Solution by Maple

Time used: 0.413 (sec). Leaf size: 31

dsolve([diff(x(t),t$2)+4*diff(x(t),t)+13*x(t)=f(t),x(0) = 0, D(x)(0) = 0],x(t), singsol=all)
 

\[ x \left (t \right ) = -\frac {\left (\int _{0}^{t}{\mathrm e}^{-2 t +2 \textit {\_U1}} \sin \left (-3 t +3 \textit {\_U1} \right ) f \left (\textit {\_U1} \right )d \textit {\_U1} \right )}{3} \]

✓ Solution by Mathematica

Time used: 0.129 (sec). Leaf size: 131

DSolve[{D[x[t],{t,2}]+4*D[x[t],t]+13*x[t]==f[t],{x[0]==0,Derivative[1][x][0] ==0}},x[t],t,IncludeSingularSolutions -> True]
 

\[ x(t)\to e^{-2 t} \left (-\sin (3 t) \int _1^0\frac {1}{3} e^{2 K[1]} \cos (3 K[1]) f(K[1])dK[1]+\sin (3 t) \int _1^t\frac {1}{3} e^{2 K[1]} \cos (3 K[1]) f(K[1])dK[1]+\cos (3 t) \left (\int _1^t-\frac {1}{3} e^{2 K[2]} f(K[2]) \sin (3 K[2])dK[2]-\int _1^0-\frac {1}{3} e^{2 K[2]} f(K[2]) \sin (3 K[2])dK[2]\right )\right ) \]