13.20 problem Problem 20

Internal problem ID [2858]
Internal file name [OUTPUT/2350_Sunday_June_05_2022_03_00_09_AM_52079326/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 20.
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]]

\[ \boxed {y^{\prime \prime }-y=8 \sin \left (t \right )-6 \cos \left (t \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 2, y^{\prime }\left (0\right ) = -1] \end {align*}

13.20.1 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 &=8 \sin \left (t \right )-6 \cos \left (t \right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-y = 8 \sin \left (t \right )-6 \cos \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 )-Y \left (s \right ) = -\frac {2 \left (-4+3 s \right )}{s^{2}+1}\tag {1} \end {align*}

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

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

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {2 s^{3}-s^{2}-4 s +7}{\left (s^{2}+1\right ) \left (s^{2}-1\right )} \end {align*}

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

Adding the above results and simplifying gives \[ y=-4 \sin \left (t \right )+3 \cos \left (t \right )+3 \sinh \left (t \right )-\cosh \left (t \right ) \] Simplifying the solution gives \[ y = -4 \sin \left (t \right )+3 \cos \left (t \right )+3 \sinh \left (t \right )-\cosh \left (t \right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -4 \sin \left (t \right )+3 \cos \left (t \right )+3 \sinh \left (t \right )-\cosh \left (t \right ) \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -4 \sin \left (t \right )+3 \cos \left (t \right )+3 \sinh \left (t \right )-\cosh \left (t \right ) \] Verified OK.

13.20.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-y=8 \sin \left (t \right )-6 \cos \left (t \right ), y \left (0\right )=2, 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 \\ {} & {} & y^{\prime \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 )=8 \sin \left (t \right )-6 \cos \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} {\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 )={\mathrm e}^{-t} \left (\int \left (-4 \sin \left (t \right )+3 \cos \left (t \right )\right ) {\mathrm e}^{t}d t \right )-{\mathrm e}^{t} \left (\int \left (-4 \sin \left (t \right )+3 \cos \left (t \right )\right ) {\mathrm e}^{-t}d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-4 \sin \left (t \right )+3 \cos \left (t \right ) \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{t}-4 \sin \left (t \right )+3 \cos \left (t \right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{t}-4 \sin \left (t \right )+3 \cos \left (t \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=c_{1} +c_{2} +3 \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{t}-4 \cos \left (t \right )-3 \sin \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=-c_{1} +c_{2} -4 \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-2, c_{2} =1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-2 \,{\mathrm e}^{-t}+{\mathrm e}^{t}-4 \sin \left (t \right )+3 \cos \left (t \right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-2 \,{\mathrm e}^{-t}+{\mathrm e}^{t}-4 \sin \left (t \right )+3 \cos \left (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: 1.953 (sec). Leaf size: 21

dsolve([diff(y(t),t$2)-y(t)=8*sin(t)-6*cos(t),y(0) = 2, D(y)(0) = -1],y(t), singsol=all)
 

\[ y \left (t \right ) = -4 \sin \left (t \right )+3 \cos \left (t \right )+3 \sinh \left (t \right )-\cosh \left (t \right ) \]

✓ Solution by Mathematica

Time used: 0.02 (sec). Leaf size: 24

DSolve[{y''[t]-y[t]==8*Sin[t]-6*Cos[t],{y[0]==2,y'[0]==-1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to -2 e^{-t}+e^t-4 \sin (t)+3 \cos (t) \]