13.25 problem Problem 25

Internal problem ID [2863]
Internal file name [OUTPUT/2355_Sunday_June_05_2022_03_00_19_AM_47489344/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 25.
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, _linear, _nonhomogeneous]]

\[ \boxed {y^{\prime \prime }+4 y=9 \sin \left (t \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = -1] \end {align*}

13.25.1 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) &=4\\ F &=9 \sin \left (t \right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+4 y = 9 \sin \left (t \right ) \end {align*}

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

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 )+4 Y \left (s \right ) = \frac {9}{s^{2}+1}\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 +4 Y \left (s \right ) = \frac {9}{s^{2}+1} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s^{3}-s^{2}+s +8}{\left (s^{2}+1\right ) \left (s^{2}+4\right )} \end {align*}

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

Adding the above results and simplifying gives \[ y=3 \sin \left (t \right )+\cos \left (2 t \right )-2 \sin \left (2 t \right ) \] Simplifying the solution gives \[ y = 3 \sin \left (t \right )+\cos \left (2 t \right )-2 \sin \left (2 t \right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 3 \sin \left (t \right )+\cos \left (2 t \right )-2 \sin \left (2 t \right ) \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = 3 \sin \left (t \right )+\cos \left (2 t \right )-2 \sin \left (2 t \right ) \] Verified OK.

13.25.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+4 y=9 \sin \left (t \right ), 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}+4=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-16}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2 \,\mathrm {I}, 2 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (2 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (2 t \right ) \\ \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} \cos \left (2 t \right )+c_{2} \sin \left (2 t \right )+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 )=9 \sin \left (t \right )\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} \cos \left (2 t \right ) & \sin \left (2 t \right ) \\ -2 \sin \left (2 t \right ) & 2 \cos \left (2 t \right ) \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 )=-9 \cos \left (2 t \right ) \left (\int \cos \left (t \right ) \sin \left (t \right )^{2}d t \right )+\frac {9 \sin \left (2 t \right ) \left (\int \left (\sin \left (3 t \right )-\sin \left (t \right )\right )d t \right )}{4} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=3 \sin \left (t \right ) \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (2 t \right )+c_{2} \sin \left (2 t \right )+3 \sin \left (t \right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (2 t \right )+c_{2} \sin \left (2 t \right )+3 \sin \left (t \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} \sin \left (2 t \right )+2 c_{2} \cos \left (2 t \right )+3 \cos \left (t \right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-1 \\ {} & {} & -1=3+2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =1, c_{2} =-2\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=3 \sin \left (t \right )+\cos \left (2 t \right )-2 \sin \left (2 t \right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=3 \sin \left (t \right )+\cos \left (2 t \right )-2 \sin \left (2 t \right ) \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: 2.047 (sec). Leaf size: 19

dsolve([diff(y(t),t$2)+4*y(t)=9*sin(t),y(0) = 1, D(y)(0) = -1],y(t), singsol=all)
 

\[ y \left (t \right ) = \cos \left (2 t \right )-2 \sin \left (2 t \right )+3 \sin \left (t \right ) \]

✓ Solution by Mathematica

Time used: 0.019 (sec). Leaf size: 20

DSolve[{y''[t]+4*y[t]==9*Sin[t],{y[0]==1,y'[0]==-1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to 3 \sin (t)-2 \sin (2 t)+\cos (2 t) \]