Internal problem ID [15566]
Internal file name [OUTPUT/15567_Tuesday_May_14_2024_10_48_02_PM_66644115/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: 844.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeff", "linear_second_order_ode_solved_by_an_integrating_factor"
Maple gives the following as the ode type
[[_2nd_order, _missing_x]]
\[ \boxed {x^{\prime \prime }+4 x^{\prime }+4 x=4} \] With initial conditions \begin {align*} [x \left (0\right ) = 1, x^{\prime }\left (0\right ) = -4] \end {align*}
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) &=4\\ q(t) &=4\\ F &=4 \end {align*}
Hence the ode is \begin {align*} x^{\prime \prime }+4 x^{\prime }+4 x = 4 \end {align*}
The domain of \(p(t)=4\) 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 )+4 s Y \left (s \right )-4 x \left (0\right )+4 Y \left (s \right ) = \frac {4}{s}\tag {1} \end {align*}
But the initial conditions are \begin {align*} x \left (0\right )&=1\\ x'(0) &=-4 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-s +4 s Y \left (s \right )+4 Y \left (s \right ) = \frac {4}{s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s^{2}+4}{s \left (s^{2}+4 s +4\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{s}-\frac {4}{\left (s +2\right )^{2}} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{s}\right ) &= 1\\ \mathcal {L}^{-1}\left (-\frac {4}{\left (s +2\right )^{2}}\right ) &= -4 t \,{\mathrm e}^{-2 t} \end {align*}
Adding the above results and simplifying gives \[ x=-4 t \,{\mathrm e}^{-2 t}+1 \] Simplifying the solution gives \[
x = -4 t \,{\mathrm e}^{-2 t}+1
\]
The solution(s) found are the following \begin{align*}
\tag{1} x &= -4 t \,{\mathrm e}^{-2 t}+1 \\
\end{align*} Verification of solutions
\[
x = -4 t \,{\mathrm e}^{-2 t}+1
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime \prime }+4 x^{\prime }+4 x=4, x \left (0\right )=1, x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-4\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}+4 r +4=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +2\right )^{2}=0 \\ \bullet & {} & \textrm {Root of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =-2 \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{1}\left (t \right )={\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} x_{1}\left (t \right )\hspace {3pt}\textrm {by}\hspace {3pt} t \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & x_{2}\left (t \right )=t \,{\mathrm e}^{-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}^{-2 t}+c_{2} {\mathrm e}^{-2 t} 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 )=4\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}^{-2 t} & t \,{\mathrm e}^{-2 t} \\ -2 \,{\mathrm e}^{-2 t} & {\mathrm e}^{-2 t}-2 t \,{\mathrm e}^{-2 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}^{-4 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} x_{p}\left (t \right ) \\ {} & {} & x_{p}\left (t \right )=-4 \,{\mathrm e}^{-2 t} \left (\int t \,{\mathrm e}^{2 t}d t -\left (\int {\mathrm e}^{2 t}d t \right ) t \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}^{-2 t}+c_{2} {\mathrm e}^{-2 t} t +1 \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} x=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-2 t} 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 }=-2 c_{1} {\mathrm e}^{-2 t}-2 c_{2} {\mathrm e}^{-2 t} t +c_{2} {\mathrm e}^{-2 t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-4 \\ {} & {} & -4=-2 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} =-4\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & x=-4 t \,{\mathrm e}^{-2 t}+1 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=-4 t \,{\mathrm e}^{-2 t}+1 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 13
\[
x = 1-4 t \,{\mathrm e}^{-2 t}
\]
✓ Solution by Mathematica
Time used: 0.014 (sec). Leaf size: 15
\[
x(t)\to 1-4 e^{-2 t} t
\]
33.15.2 Maple step by step solution
`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`
dsolve([diff(x(t),t$2)+4*diff(x(t),t)+4*x(t)=4,x(0) = 1, D(x)(0) = -4],x(t), singsol=all)
DSolve[{x''[t]+4*x'[t]+4*x[t]==4,{x[0]==1,x'[0]==-4}},x[t],t,IncludeSingularSolutions -> True]