63.9.23 problem 6

Internal problem ID [13150]
Book : A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag, NY. 2015.
Section : Chapter 2, Second order linear equations. Section 2.3.1 Nonhomogeneous Equations: Undetermined Coefficients. Exercises page 110
Problem number : 6
Date solved : Tuesday, January 28, 2025 at 05:09:24 AM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

\begin{align*} x^{\prime \prime }+2 x&=\cos \left (\sqrt {2}\, t \right ) \end{align*}

With initial conditions

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

✓ Solution by Maple

Time used: 0.018 (sec). Leaf size: 18

dsolve([diff(x(t),t$2)+2*x(t)=cos(sqrt(2)*t),x(0) = 0, D(x)(0) = 1],x(t), singsol=all)
 

\[ x \left (t \right ) = \frac {\sqrt {2}\, \sin \left (\sqrt {2}\, t \right ) \left (t +2\right )}{4} \]

✓ Solution by Mathematica

Time used: 0.144 (sec). Leaf size: 158

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

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