24.2 problem 1(b)

Internal problem ID [6500]
Internal file name [OUTPUT/5748_Sunday_June_05_2022_03_52_38_PM_12557919/index.tex]

Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 7. Laplace Transforms. Section 7.5 The Unit Step and Impulse Functions. Page 303
Problem number: 1(b).
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]]

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

24.2.1 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) &=1\\ q(t) &=-6\\ F &=t \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+y^{\prime }-6 y = t \end {align*}

The domain of \(p(t)=1\) 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 )+s Y \left (s \right )-y \left (0\right )-6 Y \left (s \right ) = \frac {1}{s^{2}}\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+s Y \left (s \right )-6 Y \left (s \right ) = \frac {1}{s^{2}} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {1}{s^{2} \left (s^{2}+s -6\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= -\frac {1}{36 s}-\frac {1}{45 \left (s +3\right )}+\frac {1}{20 s -40}-\frac {1}{6 s^{2}} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {1}{36 s}\right ) &= -{\frac {1}{36}}\\ \mathcal {L}^{-1}\left (-\frac {1}{45 \left (s +3\right )}\right ) &= -\frac {{\mathrm e}^{-3 t}}{45}\\ \mathcal {L}^{-1}\left (\frac {1}{20 s -40}\right ) &= \frac {{\mathrm e}^{2 t}}{20}\\ \mathcal {L}^{-1}\left (-\frac {1}{6 s^{2}}\right ) &= -\frac {t}{6} \end {align*}

Adding the above results and simplifying gives \[ y=-\frac {1}{36}-\frac {t}{6}+\frac {{\mathrm e}^{2 t}}{20}-\frac {{\mathrm e}^{-3 t}}{45} \] Simplifying the solution gives \[ y = -\frac {\left (-9 \,{\mathrm e}^{5 t}+30 t \,{\mathrm e}^{3 t}+5 \,{\mathrm e}^{3 t}+4\right ) {\mathrm e}^{-3 t}}{180} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (-9 \,{\mathrm e}^{5 t}+30 t \,{\mathrm e}^{3 t}+5 \,{\mathrm e}^{3 t}+4\right ) {\mathrm e}^{-3 t}}{180} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -\frac {\left (-9 \,{\mathrm e}^{5 t}+30 t \,{\mathrm e}^{3 t}+5 \,{\mathrm e}^{3 t}+4\right ) {\mathrm e}^{-3 t}}{180} \] Verified OK.

24.2.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+y^{\prime }-6 y=t , y \left (0\right )=0, y^{\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 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+r -6=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +3\right ) \left (r -2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3, 2\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-3 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{2 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}^{-3 t}+c_{2} {\mathrm e}^{2 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 )=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}^{-3 t} & {\mathrm e}^{2 t} \\ -3 \,{\mathrm e}^{-3 t} & 2 \,{\mathrm e}^{2 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=5 \,{\mathrm e}^{-t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=\frac {\left ({\mathrm e}^{5 t} \left (\int t \,{\mathrm e}^{-2 t}d t \right )-\left (\int t \,{\mathrm e}^{3 t}d t \right )\right ) {\mathrm e}^{-3 t}}{5} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {1}{36}-\frac {t}{6} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-3 t}+c_{2} {\mathrm e}^{2 t}-\frac {1}{36}-\frac {t}{6} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-3 t}+c_{2} {\mathrm e}^{2 t}-\frac {1}{36}-\frac {t}{6} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} -\frac {1}{36} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 c_{1} {\mathrm e}^{-3 t}+2 c_{2} {\mathrm e}^{2 t}-\frac {1}{6} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-3 c_{1} +2 c_{2} -\frac {1}{6} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {1}{45}, c_{2} =\frac {1}{20}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {\left (-9 \,{\mathrm e}^{5 t}+30 t \,{\mathrm e}^{3 t}+5 \,{\mathrm e}^{3 t}+4\right ) {\mathrm e}^{-3 t}}{180} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {\left (-9 \,{\mathrm e}^{5 t}+30 t \,{\mathrm e}^{3 t}+5 \,{\mathrm e}^{3 t}+4\right ) {\mathrm e}^{-3 t}}{180} \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: 1.531 (sec). Leaf size: 21

dsolve([diff(y(t),t$2)+diff(y(t),t)-6*y(t)=t,y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {\left (9 \,{\mathrm e}^{5 t}-30 t \,{\mathrm e}^{3 t}-5 \,{\mathrm e}^{3 t}-4\right ) {\mathrm e}^{-3 t}}{180} \]

✓ Solution by Mathematica

Time used: 0.015 (sec). Leaf size: 28

DSolve[{y''[t]+y'[t]-6*y[t]==t,{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{180} \left (-30 t-4 e^{-3 t}+9 e^{2 t}-5\right ) \]