Internal problem ID [1859]
Book: Differential equations and their applications, 4th ed., M. Braun
Section: Section 3.12, Systems of differential equations. The nonhomogeneous equation. variation
of parameters. Page 366
Problem number: 4.
ODE order: 1.
ODE degree: 1.
Solve \begin {align*} x_{1}^{\prime }\left (t \right )&=x_{2} \left (t \right )+f_{1} \left (t \right )\\ x_{2}^{\prime }\left (t \right )&=-x_{1} \left (t \right )+f_{2} \left (t \right ) \end {align*}
With initial conditions \[ [x_{1} \left (0\right ) = 0, x_{2} \left (0\right ) = 0] \]
✓ Solution by Maple
Time used: 0.187 (sec). Leaf size: 101
dsolve([diff(x__1(t),t) = x__2(t)+f__1(t), diff(x__2(t),t) = -x__1(t)+f__2(t), x__1(0) = 0, x__2(0) = 0], singsol=all)
\begin{align*} x_{1} \left (t \right ) &= f_{1} \left (0\right ) \sin \left (t \right )+\left (\int _{0}^{t}\cos \left (\textit {\_z1} \right ) \left (\frac {d}{d \textit {\_z1}}f_{1} \left (\textit {\_z1} \right )+f_{2} \left (\textit {\_z1} \right )\right )d \textit {\_z1} \right ) \sin \left (t \right )-\left (\int _{0}^{t}\sin \left (\textit {\_z1} \right ) \left (\frac {d}{d \textit {\_z1}}f_{1} \left (\textit {\_z1} \right )+f_{2} \left (\textit {\_z1} \right )\right )d \textit {\_z1} \right ) \cos \left (t \right ) \\ x_{2} \left (t \right ) &= f_{1} \left (0\right ) \cos \left (t \right )+\left (\int _{0}^{t}\cos \left (\textit {\_z1} \right ) \left (\frac {d}{d \textit {\_z1}}f_{1} \left (\textit {\_z1} \right )+f_{2} \left (\textit {\_z1} \right )\right )d \textit {\_z1} \right ) \cos \left (t \right )+\left (\int _{0}^{t}\sin \left (\textit {\_z1} \right ) \left (\frac {d}{d \textit {\_z1}}f_{1} \left (\textit {\_z1} \right )+f_{2} \left (\textit {\_z1} \right )\right )d \textit {\_z1} \right ) \sin \left (t \right )-f_{1} \left (t \right ) \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.035 (sec). Leaf size: 216
DSolve[{x1'[t]==0*x1[t]+1*x2[t]+f1[t],x2'[t]==-1*x1[t]-0*x2[t]+f2[t]},{x1[0]==0,x2[0]==0},{x1[t],x2[t]},t,IncludeSingularSolutions -> True]
\begin{align*} \text {x1}(t)\to -\cos (t) \int _1^0(\cos (K[1]) \text {f1}(K[1])-\text {f2}(K[1]) \sin (K[1]))dK[1]+\cos (t) \int _1^t(\cos (K[1]) \text {f1}(K[1])-\text {f2}(K[1]) \sin (K[1]))dK[1]+\sin (t) \left (\int _1^t(\cos (K[2]) \text {f2}(K[2])+\text {f1}(K[2]) \sin (K[2]))dK[2]-\int _1^0(\cos (K[2]) \text {f2}(K[2])+\text {f1}(K[2]) \sin (K[2]))dK[2]\right ) \\ \text {x2}(t)\to \sin (t) \int _1^0(\cos (K[1]) \text {f1}(K[1])-\text {f2}(K[1]) \sin (K[1]))dK[1]-\sin (t) \int _1^t(\cos (K[1]) \text {f1}(K[1])-\text {f2}(K[1]) \sin (K[1]))dK[1]+\cos (t) \left (\int _1^t(\cos (K[2]) \text {f2}(K[2])+\text {f1}(K[2]) \sin (K[2]))dK[2]-\int _1^0(\cos (K[2]) \text {f2}(K[2])+\text {f1}(K[2]) \sin (K[2]))dK[2]\right ) \\ \end{align*}