problem 8

Internal problem ID [11522]
Internal ﬁle name [OUTPUT/10505_Thursday_May_18_2023_04_21_23_AM_33382183/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 3.3 The convolution property. Exercises page 162
Problem number: 8.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {x^{\prime \prime }+3 x^{\prime }+2 x={\mathrm e}^{-4 t}} \] With initial conditions \begin {align*} [x \left (0\right ) = 0, x^{\prime }\left (0\right ) = 0] \end {align*}

16.2.1 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) &=3\\ q(t) &=2\\ F &={\mathrm e}^{-4 t} \end {align*}

Hence the ode is \begin {align*} x^{\prime \prime }+3 x^{\prime }+2 x = {\mathrm e}^{-4 t} \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 )+3 s Y \left (s \right )-3 x \left (0\right )+2 Y \left (s \right ) = \frac {1}{s +4}\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+3 s Y \left (s \right )+2 Y \left (s \right ) = \frac {1}{s +4} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {1}{\left (s +4\right ) \left (s^{2}+3 s +2\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= -\frac {1}{2 \left (s +2\right )}+\frac {1}{3 s +3}+\frac {1}{6 s +24} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {1}{2 \left (s +2\right )}\right ) &= -\frac {{\mathrm e}^{-2 t}}{2}\\ \mathcal {L}^{-1}\left (\frac {1}{3 s +3}\right ) &= \frac {{\mathrm e}^{-t}}{3}\\ \mathcal {L}^{-1}\left (\frac {1}{6 s +24}\right ) &= \frac {{\mathrm e}^{-4 t}}{6} \end {align*}

Adding the above results and simplifying gives \[ x=\frac {{\mathrm e}^{-t}}{3}-\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-4 t}}{6} \] Simplifying the solution gives \[ x = \frac {{\mathrm e}^{-t}}{3}-\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-4 t}}{6} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} x &= \frac {{\mathrm e}^{-t}}{3}-\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-4 t}}{6} \\ \end{align*}

Veriﬁcation of solutions

\[ x = \frac {{\mathrm e}^{-t}}{3}-\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-4 t}}{6} \] Veriﬁed OK.

16.2.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}x^{\prime }+3 x^{\prime }+2 x={\mathrm e}^{-4 t}, x \left (0\right )=0, 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 \\ {} & {} & \frac {d}{d t}x^{\prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+3 r +2=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +2\right ) \left (r +1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2, -1\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{1}\left (t \right )={\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{2}\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}^{-2 t}+c_{2} {\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 )={\mathrm e}^{-4 t}\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} & {\mathrm e}^{-t} \\ -2 \,{\mathrm e}^{-2 t} & -{\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}^{-3 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} x_{p}\left (t \right ) \\ {} & {} & x_{p}\left (t \right )=-{\mathrm e}^{-2 t} \left (\int {\mathrm e}^{-2 t}d t \right )+{\mathrm e}^{-t} \left (\int {\mathrm e}^{-3 t}d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & x_{p}\left (t \right )=\frac {{\mathrm e}^{-4 t}}{6} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & x=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-t}+\frac {{\mathrm e}^{-4 t}}{6} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} x=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-t}+\frac {{\mathrm e}^{-4 t}}{6} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} +\frac {1}{6} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & x^{\prime }=-2 c_{1} {\mathrm e}^{-2 t}-c_{2} {\mathrm e}^{-t}-\frac {2 \,{\mathrm e}^{-4 t}}{3} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-2 c_{1} -c_{2} -\frac {2}{3} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {1}{2}, c_{2} =\frac {1}{3}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & x=\frac {{\mathrm e}^{-t}}{3}-\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-4 t}}{6} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=\frac {{\mathrm e}^{-t}}{3}-\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-4 t}}{6} \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)] 
<- double symmetry of the form [xi=0, eta=F(x)] successful`

\[ x \left (t \right ) = \frac {{\mathrm e}^{-4 t}}{6}+\frac {{\mathrm e}^{-t}}{3}-\frac {{\mathrm e}^{-2 t}}{2} \]

\[ x(t)\to \frac {1}{6} e^{-4 t} \left (e^t-1\right )^2 \left (2 e^t+1\right ) \]

16.2 problem 8

16.2.1 Existence and uniqueness analysis

16.2.2 Maple step by step solution