15.9 problem 6(i)

Internal problem ID [11515]
Internal file name [OUTPUT/10498_Thursday_May_18_2023_04_21_10_AM_77327639/index.tex]

Book: A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag, NY. 2015.
Section: Chapter 3, Laplace transform. Section 3.2.1 Initial value problems. Exercises page 156
Problem number: 6(i).
ODE order: 2.
ODE degree: 1.

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

Maple gives the following as the ode type

[[_2nd_order, _missing_x]]

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

15.9.1 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) &=-2\\ F &=1 \end {align*}

Hence the ode is \begin {align*} x^{\prime \prime }-2 x = 1 \end {align*}

The domain of \(p(t)=0\) is \[ \{-\infty

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 )-2 Y \left (s \right ) = \frac {1}{s}\tag {1} \end {align*}

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

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

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

Applying partial fractions decomposition results in \[ Y(s)= \frac {3}{4 \left (s -\sqrt {2}\right )}+\frac {3}{4 \left (s +\sqrt {2}\right )}-\frac {1}{2 s} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {3}{4 \left (s -\sqrt {2}\right )}\right ) &= \frac {3 \,{\mathrm e}^{\sqrt {2}\, t}}{4}\\ \mathcal {L}^{-1}\left (\frac {3}{4 \left (s +\sqrt {2}\right )}\right ) &= \frac {3 \,{\mathrm e}^{-\sqrt {2}\, t}}{4}\\ \mathcal {L}^{-1}\left (-\frac {1}{2 s}\right ) &= -{\frac {1}{2}} \end {align*}

Adding the above results and simplifying gives \[ x=-\frac {1}{2}+\frac {3 \cosh \left (\sqrt {2}\, t \right )}{2} \] Simplifying the solution gives \[ x = -\frac {1}{2}+\frac {3 \cosh \left (\sqrt {2}\, t \right )}{2} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} x &= -\frac {1}{2}+\frac {3 \cosh \left (\sqrt {2}\, t \right )}{2} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ x = -\frac {1}{2}+\frac {3 \cosh \left (\sqrt {2}\, t \right )}{2} \] Verified OK.

15.9.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime \prime }-2 x=1, x \left (0\right )=1, x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0\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}-2=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {8}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (\sqrt {2}, -\sqrt {2}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{1}\left (t \right )={\mathrm e}^{\sqrt {2}\, t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{2}\left (t \right )={\mathrm e}^{-\sqrt {2}\, t} \\ \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} {\mathrm e}^{\sqrt {2}\, t}+c_{2} {\mathrm e}^{-\sqrt {2}\, t}+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 )=1\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} {\mathrm e}^{\sqrt {2}\, t} & {\mathrm e}^{-\sqrt {2}\, t} \\ \sqrt {2}\, {\mathrm e}^{\sqrt {2}\, t} & -\sqrt {2}\, {\mathrm e}^{-\sqrt {2}\, t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )=-2 \sqrt {2} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} x_{p}\left (t \right ) \\ {} & {} & x_{p}\left (t \right )=\frac {\sqrt {2}\, \left ({\mathrm e}^{\sqrt {2}\, t} \left (\int {\mathrm e}^{-\sqrt {2}\, t}d t \right )-{\mathrm e}^{-\sqrt {2}\, t} \left (\int {\mathrm e}^{\sqrt {2}\, t}d t \right )\right )}{4} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & x_{p}\left (t \right )=-\frac {1}{2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & x=c_{1} {\mathrm e}^{\sqrt {2}\, t}+c_{2} {\mathrm e}^{-\sqrt {2}\, t}-\frac {1}{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} x=c_{1} {\mathrm e}^{\sqrt {2}\, t}+c_{2} {\mathrm e}^{-\sqrt {2}\, t}-\frac {1}{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=1 \\ {} & {} & 1=c_{1} +c_{2} -\frac {1}{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & x^{\prime }=c_{1} \sqrt {2}\, {\mathrm e}^{\sqrt {2}\, t}-c_{2} \sqrt {2}\, {\mathrm e}^{-\sqrt {2}\, t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{1} \sqrt {2}-\sqrt {2}\, c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {3}{4}, c_{2} =\frac {3}{4}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & x=\frac {3 \,{\mathrm e}^{\sqrt {2}\, t}}{4}+\frac {3 \,{\mathrm e}^{-\sqrt {2}\, t}}{4}-\frac {1}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=\frac {3 \,{\mathrm e}^{\sqrt {2}\, t}}{4}+\frac {3 \,{\mathrm e}^{-\sqrt {2}\, t}}{4}-\frac {1}{2} \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`
 

✓ Solution by Maple

Time used: 4.172 (sec). Leaf size: 14

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

\[ x \left (t \right ) = -\frac {1}{2}+\frac {3 \cosh \left (\sqrt {2}\, t \right )}{2} \]

✓ Solution by Mathematica

Time used: 0.029 (sec). Leaf size: 34

DSolve[{x''[t]-2*x[t]==1,{x[0]==1,x'[0]==0}},x[t],t,IncludeSingularSolutions -> True]
 

\[ x(t)\to \frac {1}{4} \left (3 e^{-\sqrt {2} t}+3 e^{\sqrt {2} t}-2\right ) \]