6.10 problem 30

Internal problem ID [6677]
Internal file name [OUTPUT/5925_Sunday_June_05_2022_04_01_50_PM_31220714/index.tex]

Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section: CHAPTER 7 THE LAPLACE TRANSFORM. 7.3.1 TRANSLATION ON THE s-AXIS. Page 297
Problem number: 30.
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, _with_linear_symmetries]]

\[ \boxed {y^{\prime \prime }-2 y^{\prime }+5 y=t +1} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 4] \end {align*}

6.10.1 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) &=-2\\ q(t) &=5\\ F &=t +1 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-2 y^{\prime }+5 y = t +1 \end {align*}

The domain of \(p(t)=-2\) 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 )-2 s Y \left (s \right )+2 y \left (0\right )+5 Y \left (s \right ) = \frac {s +1}{s^{2}}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y'(0) &=4 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-4-2 s Y \left (s \right )+5 Y \left (s \right ) = \frac {s +1}{s^{2}} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {4 s^{2}+s +1}{s^{2} \left (s^{2}-2 s +5\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {-\frac {7}{50}-\frac {51 i}{50}}{s -1-2 i}+\frac {-\frac {7}{50}+\frac {51 i}{50}}{s -1+2 i}+\frac {1}{5 s^{2}}+\frac {7}{25 s} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {-\frac {7}{50}-\frac {51 i}{50}}{s -1-2 i}\right ) &= \left (-\frac {7}{50}-\frac {51 i}{50}\right ) {\mathrm e}^{\left (1+2 i\right ) t}\\ \mathcal {L}^{-1}\left (\frac {-\frac {7}{50}+\frac {51 i}{50}}{s -1+2 i}\right ) &= \left (-\frac {7}{50}+\frac {51 i}{50}\right ) {\mathrm e}^{\left (1-2 i\right ) t}\\ \mathcal {L}^{-1}\left (\frac {1}{5 s^{2}}\right ) &= \frac {t}{5}\\ \mathcal {L}^{-1}\left (\frac {7}{25 s}\right ) &= {\frac {7}{25}} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {7}{25}+\frac {t}{5}+\frac {\left (-7 \cos \left (2 t \right )+51 \sin \left (2 t \right )\right ) {\mathrm e}^{t}}{25} \] Simplifying the solution gives \[ y = -\frac {7 \,{\mathrm e}^{t} \cos \left (2 t \right )}{25}+\frac {51 \,{\mathrm e}^{t} \sin \left (2 t \right )}{25}+\frac {t}{5}+\frac {7}{25} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {7 \,{\mathrm e}^{t} \cos \left (2 t \right )}{25}+\frac {51 \,{\mathrm e}^{t} \sin \left (2 t \right )}{25}+\frac {t}{5}+\frac {7}{25} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -\frac {7 \,{\mathrm e}^{t} \cos \left (2 t \right )}{25}+\frac {51 \,{\mathrm e}^{t} \sin \left (2 t \right )}{25}+\frac {t}{5}+\frac {7}{25} \] Verified OK.

6.10.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }-2 y^{\prime }+5 y=t +1, y \left (0\right )=0, y^{\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 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}-2 r +5=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {2\pm \left (\sqrt {-16}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (1-2 \,\mathrm {I}, 1+2 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{t} \cos \left (2 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{t} \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} {\mathrm e}^{t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{t} \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 )=t +1\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} \cos \left (2 t \right ) & {\mathrm e}^{t} \sin \left (2 t \right ) \\ {\mathrm e}^{t} \cos \left (2 t \right )-2 \,{\mathrm e}^{t} \sin \left (2 t \right ) & {\mathrm e}^{t} \sin \left (2 t \right )+2 \,{\mathrm e}^{t} \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 \,{\mathrm e}^{2 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {{\mathrm e}^{t} \left (\cos \left (2 t \right ) \left (\int \left (t +1\right ) \sin \left (2 t \right ) {\mathrm e}^{-t}d t \right )-\sin \left (2 t \right ) \left (\int \left (t +1\right ) \cos \left (2 t \right ) {\mathrm e}^{-t}d t \right )\right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {7}{25}+\frac {t}{5} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{t} \sin \left (2 t \right )+\frac {7}{25}+\frac {t}{5} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{t} \sin \left (2 t \right )+\frac {7}{25}+\frac {t}{5} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +\frac {7}{25} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{1} {\mathrm e}^{t} \cos \left (2 t \right )-2 c_{1} {\mathrm e}^{t} \sin \left (2 t \right )+c_{2} {\mathrm e}^{t} \sin \left (2 t \right )+2 c_{2} {\mathrm e}^{t} \cos \left (2 t \right )+\frac {1}{5} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=4 \\ {} & {} & 4=c_{1} +\frac {1}{5}+2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {7}{25}, c_{2} =\frac {51}{25}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {7 \,{\mathrm e}^{t} \cos \left (2 t \right )}{25}+\frac {51 \,{\mathrm e}^{t} \sin \left (2 t \right )}{25}+\frac {t}{5}+\frac {7}{25} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {7 \,{\mathrm e}^{t} \cos \left (2 t \right )}{25}+\frac {51 \,{\mathrm e}^{t} \sin \left (2 t \right )}{25}+\frac {t}{5}+\frac {7}{25} \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: 1.797 (sec). Leaf size: 26

dsolve([diff(y(t),t$2)-2*diff(y(t),t)+5*y(t)=1+t,y(0) = 0, D(y)(0) = 4],y(t), singsol=all)
 

\[ y \left (t \right ) = -\frac {7 \,{\mathrm e}^{t} \cos \left (2 t \right )}{25}+\frac {51 \,{\mathrm e}^{t} \sin \left (2 t \right )}{25}+\frac {t}{5}+\frac {7}{25} \]

✓ Solution by Mathematica

Time used: 0.017 (sec). Leaf size: 32

DSolve[{y''[t]-2*y'[t]+5*y[t]==1+t,{y[0]==0,y'[0]==4}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{25} \left (5 t+51 e^t \sin (2 t)-7 e^t \cos (2 t)+7\right ) \]