problem 846

Internal problem ID [15568]
Internal ﬁle name [OUTPUT/15569_Tuesday_May_14_2024_10_48_03_PM_85677221/index.tex]

Book: A book of problems in ordinary diﬀerential equations. M.L. KRASNOV, A.L. KISELYOV, G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 3. Section 24.2. Solving the Cauchy problem for linear diﬀerential equation with constant coeﬃcients. Exercises page 249
Problem number: 846.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeﬀ"

\[ \boxed {x^{\prime \prime }+x=2 \cos \left (t \right )} \] With initial conditions \begin {align*} [x \left (0\right ) = -1, x^{\prime }\left (0\right ) = 1] \end {align*}

33.17.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} x^{\prime \prime } + p(t)x^{\prime } + q(t) x &= F \end {align*}

Where here \begin {align*} p(t) &=0\\ q(t) &=1\\ F &=2 \cos \left (t \right ) \end {align*}

Hence the ode is \begin {align*} x^{\prime \prime }+x = 2 \cos \left (t \right ) \end {align*}

Solving using the Laplace transform method. Let \begin {align*} \mathcal {L}\left (x\right ) =Y(s) \end {align*}

Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (x^{\prime }\right ) &= s Y(s) - x \left (0\right )\\ \mathcal {L}\left (x^{\prime \prime }\right ) &= s^2 Y(s) - x'(0) - s x \left (0\right ) \end {align*}

The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s^{2} Y \left (s \right )-x^{\prime }\left (0\right )-s x \left (0\right )+Y \left (s \right ) = \frac {2 s}{s^{2}+1}\tag {1} \end {align*}

But the initial conditions are \begin {align*} x \left (0\right )&=-1\\ x'(0) &=1 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-1+s +Y \left (s \right ) = \frac {2 s}{s^{2}+1} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {s^{3}-s^{2}-s -1}{\left (s^{2}+1\right )^{2}} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= -\frac {i}{2 \left (s -i\right )^{2}}+\frac {i}{2 \left (s +i\right )^{2}}+\frac {-\frac {1}{2}-\frac {i}{2}}{s -i}+\frac {-\frac {1}{2}+\frac {i}{2}}{s +i} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {i}{2 \left (s -i\right )^{2}}\right ) &= -\frac {i t \,{\mathrm e}^{i t}}{2}\\ \mathcal {L}^{-1}\left (\frac {i}{2 \left (s +i\right )^{2}}\right ) &= \frac {i t \,{\mathrm e}^{-i t}}{2}\\ \mathcal {L}^{-1}\left (\frac {-\frac {1}{2}-\frac {i}{2}}{s -i}\right ) &= \left (-\frac {1}{2}-\frac {i}{2}\right ) {\mathrm e}^{i t}\\ \mathcal {L}^{-1}\left (\frac {-\frac {1}{2}+\frac {i}{2}}{s +i}\right ) &= \left (-\frac {1}{2}+\frac {i}{2}\right ) {\mathrm e}^{-i t} \end {align*}

Adding the above results and simplifying gives \[ x=-\cos \left (t \right )+\left (t +1\right ) \sin \left (t \right ) \] Simplifying the solution gives \[ x = -\cos \left (t \right )+\left (t +1\right ) \sin \left (t \right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} x &= -\cos \left (t \right )+\left (t +1\right ) \sin \left (t \right ) \\ \end{align*}

Veriﬁcation of solutions

\[ x = -\cos \left (t \right )+\left (t +1\right ) \sin \left (t \right ) \] Veriﬁed OK.

33.17.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime \prime }+x=2 \cos \left (t \right ), x \left (0\right )=-1, x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & x^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+1=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (\mathrm {-I}, \mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{1}\left (t \right )=\cos \left (t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{2}\left (t \right )=\sin \left (t \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & x=c_{1} x_{1}\left (t \right )+c_{2} x_{2}\left (t \right )+x_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x=c_{1} \cos \left (t \right )+c_{2} \sin \left (t \right )+x_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} x_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} x_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [x_{p}\left (t \right )=-x_{1}\left (t \right ) \left (\int \frac {x_{2}\left (t \right ) f \left (t \right )}{W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )}d t \right )+x_{2}\left (t \right ) \left (\int \frac {x_{1}\left (t \right ) f \left (t \right )}{W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )}d t \right ), f \left (t \right )=2 \cos \left (t \right )\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )=\left [\begin {array}{cc} \cos \left (t \right ) & \sin \left (t \right ) \\ -\sin \left (t \right ) & \cos \left (t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )=1 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} x_{p}\left (t \right ) \\ {} & {} & x_{p}\left (t \right )=-\cos \left (t \right ) \left (\int \sin \left (2 t \right )d t \right )+2 \sin \left (t \right ) \left (\int \cos \left (t \right )^{2}d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & x_{p}\left (t \right )=\frac {\cos \left (t \right )}{2}+\sin \left (t \right ) t \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & x=c_{1} \cos \left (t \right )+c_{2} \sin \left (t \right )+\frac {\cos \left (t \right )}{2}+\sin \left (t \right ) t \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} x=c_{1} \cos \left (t \right )+c_{2} \sin \left (t \right )+\frac {\cos \left (t \right )}{2}+\sin \left (t \right ) t \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=-1 \\ {} & {} & -1=\frac {1}{2}+c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & x^{\prime }=-c_{1} \sin \left (t \right )+c_{2} \cos \left (t \right )+\frac {\sin \left (t \right )}{2}+t \cos \left (t \right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {3}{2}, c_{2} =1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & x=\sin \left (t \right ) t -\cos \left (t \right )+\sin \left (t \right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=\sin \left (t \right ) t -\cos \left (t \right )+\sin \left (t \right ) \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   checking if the LODE has constant coefficients 
   <- constant coefficients successful 
<- solving first the homogeneous part of the ODE successful`

33.17 problem 846

33.17.1 Existence and uniqueness analysis

33.17.2 Maple step by step solution