problem Problem 15

Internal problem ID [2853]
Internal ﬁle name [OUTPUT/2345_Sunday_June_05_2022_02_59_59_AM_98155230/index.tex]

Book: Diﬀerential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 10, The Laplace Transform and Some Elementary Applications. Exercises for 10.4. page 689
Problem number: Problem 15.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }-y=12 \,{\mathrm e}^{2 t}} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 1] \end {align*}

13.15.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(t) &=0\\ q(t) &=-1\\ F &=12 \,{\mathrm e}^{2 t} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-y = 12 \,{\mathrm e}^{2 t} \end {align*}

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

Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (y^{\prime }\right ) &= s Y(s) - y \left (0\right )\\ \mathcal {L}\left (y^{\prime \prime }\right ) &= s^2 Y(s) - y'(0) - s y \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 )-y^{\prime }\left (0\right )-s y \left (0\right )-Y \left (s \right ) = \frac {12}{s -2}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=1\\ y'(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 {12}{s -2} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s^{2}-s +10}{\left (s -2\right ) \left (s^{2}-1\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {4}{s -2}+\frac {2}{s +1}-\frac {5}{s -1} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {4}{s -2}\right ) &= 4 \,{\mathrm e}^{2 t}\\ \mathcal {L}^{-1}\left (\frac {2}{s +1}\right ) &= 2 \,{\mathrm e}^{-t}\\ \mathcal {L}^{-1}\left (-\frac {5}{s -1}\right ) &= -5 \,{\mathrm e}^{t} \end {align*}

Adding the above results and simplifying gives \[ y=-3 \cosh \left (t \right )-7 \sinh \left (t \right )+4 \,{\mathrm e}^{2 t} \] Simplifying the solution gives \[ y = -3 \cosh \left (t \right )-7 \sinh \left (t \right )+4 \,{\mathrm e}^{2 t} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -3 \cosh \left (t \right )-7 \sinh \left (t \right )+4 \,{\mathrm e}^{2 t} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -3 \cosh \left (t \right )-7 \sinh \left (t \right )+4 \,{\mathrm e}^{2 t} \] Veriﬁed OK.

13.15.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }-y=12 \,{\mathrm e}^{2 t}, y \left (0\right )=1, y^{\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 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}-1=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -1\right ) \left (r +1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1, 1\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right )+y_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{t}+y_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} y_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (t \right )=-y_{1}\left (t \right ) \left (\int \frac {y_{2}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right )+y_{2}\left (t \right ) \left (\int \frac {y_{1}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right ), f \left (t \right )=12 \,{\mathrm e}^{2 t}\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\left [\begin {array}{cc} {\mathrm e}^{-t} & {\mathrm e}^{t} \\ -{\mathrm e}^{-t} & {\mathrm e}^{t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=2 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-6 \,{\mathrm e}^{-t} \left (\int {\mathrm e}^{3 t}d t \right )+6 \,{\mathrm e}^{t} \left (\int {\mathrm e}^{t}d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=4 \,{\mathrm e}^{2 t} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{t}+4 \,{\mathrm e}^{2 t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{t}+4 {\mathrm e}^{2 t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} +c_{2} +4 \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{t}+8 \,{\mathrm e}^{2 t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-c_{1} +c_{2} +8 \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =2, c_{2} =-5\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=2 \,{\mathrm e}^{-t}-5 \,{\mathrm e}^{t}+4 \,{\mathrm e}^{2 t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=2 \,{\mathrm e}^{-t}-5 \,{\mathrm e}^{t}+4 \,{\mathrm e}^{2 t} \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`

\[ y \left (t \right ) = -3 \cosh \left (t \right )-7 \sinh \left (t \right )+4 \,{\mathrm e}^{2 t} \]

13.15 problem Problem 15

13.15.1 Existence and uniqueness analysis

13.15.2 Maple step by step solution