33.14 problem 843

Internal problem ID [15565]
Internal file name [OUTPUT/15566_Tuesday_May_14_2024_10_48_02_PM_31591721/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: 843.
ODE order: 2.
ODE degree: 1.

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

Maple gives the following as the ode type

[[_2nd_order, _missing_x]]

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

33.14.1 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) &=-2\\ q(t) &=2\\ F &=2 \end {align*}

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

The domain of \(p(t)=-2\) 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 s Y \left (s \right )+2 x \left (0\right )+2 Y \left (s \right ) = \frac {2}{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 )+2-s -2 s Y \left (s \right )+2 Y \left (s \right ) = \frac {2}{s} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {1}{s} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} x&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {1}{s}\right )\\ &= 1 \end {align*}

Simplifying the solution gives \[ x = 1 \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} x &= 1 \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ x = 1 \] Verified OK.

33.14.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime \prime }-2 x^{\prime }+2 x=2, 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 r +2=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {2\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (1-\mathrm {I}, 1+\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{1}\left (t \right )={\mathrm e}^{t} \cos \left (t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{2}\left (t \right )=\sin \left (t \right ) {\mathrm e}^{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}^{t} \cos \left (t \right )+c_{2} \sin \left (t \right ) {\mathrm e}^{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 )=2\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}^{t} \cos \left (t \right ) & \sin \left (t \right ) {\mathrm e}^{t} \\ {\mathrm e}^{t} \cos \left (t \right )-\sin \left (t \right ) {\mathrm e}^{t} & {\mathrm e}^{t} \cos \left (t \right )+\sin \left (t \right ) {\mathrm e}^{t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )={\mathrm e}^{2 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} x_{p}\left (t \right ) \\ {} & {} & x_{p}\left (t \right )=-2 \,{\mathrm e}^{t} \left (\cos \left (t \right ) \left (\int {\mathrm e}^{-t} \sin \left (t \right )d t \right )-\sin \left (t \right ) \left (\int {\mathrm e}^{-t} \cos \left (t \right )d t \right )\right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & x_{p}\left (t \right )=1 \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & x=c_{1} {\mathrm e}^{t} \cos \left (t \right )+c_{2} \sin \left (t \right ) {\mathrm e}^{t}+1 \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} x=c_{1} {\mathrm e}^{t} \cos \left (t \right )+c_{2} \sin \left (t \right ) {\mathrm e}^{t}+1 \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=1 \\ {} & {} & 1=c_{1} +1 \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & x^{\prime }=c_{1} {\mathrm e}^{t} \cos \left (t \right )-c_{1} {\mathrm e}^{t} \sin \left (t \right )+c_{2} \cos \left (t \right ) {\mathrm e}^{t}+c_{2} \sin \left (t \right ) {\mathrm e}^{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} +c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =0\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & x=1 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=1 \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: 0.016 (sec). Leaf size: 5

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

\[ x = 1 \]

✓ Solution by Mathematica

Time used: 0.023 (sec). Leaf size: 6

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

\[ x(t)\to 1 \]