problem 68

Internal problem ID [6685]
Internal ﬁle name [OUTPUT/5933_Sunday_June_05_2022_04_02_20_PM_64677786/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: 68.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }-5 y^{\prime }+6 y=\operatorname {Heaviside}\left (t -1\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}

6.18.1 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) &=-5\\ q(t) &=6\\ F &=\operatorname {Heaviside}\left (t -1\right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-5 y^{\prime }+6 y = \operatorname {Heaviside}\left (t -1\right ) \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 )-5 s Y \left (s \right )+5 y \left (0\right )+6 Y \left (s \right ) = \frac {{\mathrm e}^{-s}}{s}\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-1-5 s Y \left (s \right )+6 Y \left (s \right ) = \frac {{\mathrm e}^{-s}}{s} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {{\mathrm e}^{-s}+s}{s \left (s^{2}-5 s +6\right )} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {{\mathrm e}^{-s}+s}{s \left (s^{2}-5 s +6\right )}\right )\\ &= \frac {\operatorname {Heaviside}\left (t -1\right )}{6}+{\mathrm e}^{3 t}-{\mathrm e}^{2 t}+\frac {{\mathrm e}^{3 t -3}}{3}-\frac {{\mathrm e}^{2 t -2}}{2}+\frac {\left (-2 \,{\mathrm e}^{3 t -3}+3 \,{\mathrm e}^{2 t -2}\right ) \operatorname {Heaviside}\left (1-t \right )}{6} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} {\mathrm e}^{3 t}-{\mathrm e}^{2 t} & t <1 \\ {\mathrm e}^{3}-{\mathrm e}^{2}+\frac {1}{6} & t =1 \\ {\mathrm e}^{3 t}-{\mathrm e}^{2 t}+\frac {{\mathrm e}^{3 t -3}}{3}-\frac {{\mathrm e}^{2 t -2}}{2}+\frac {1}{6} & 1

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left \{\begin {array}{cc} {\mathrm e}^{3 t}-{\mathrm e}^{2 t} & t <1 \\ {\mathrm e}^{3}-{\mathrm e}^{2}+\frac {1}{6} & t &=1 \\ {\mathrm e}^{3 t}-{\mathrm e}^{2 t}+\frac {{\mathrm e}^{3 t -3}}{3}-\frac {{\mathrm e}^{2 t -2}}{2}+\frac {1}{6} & 1

Veriﬁcation of solutions

\[ y = \left \{\begin {array}{cc} {\mathrm e}^{3 t}-{\mathrm e}^{2 t} & t <1 \\ {\mathrm e}^{3}-{\mathrm e}^{2}+\frac {1}{6} & t =1 \\ {\mathrm e}^{3 t}-{\mathrm e}^{2 t}+\frac {{\mathrm e}^{3 t -3}}{3}-\frac {{\mathrm e}^{2 t -2}}{2}+\frac {1}{6} & 1

6.18.2 Maple step by step solution

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

\[ y \left (t \right ) = \frac {\operatorname {Heaviside}\left (t -1\right )}{6}-{\mathrm e}^{2 t}+{\mathrm e}^{3 t}-\frac {\operatorname {Heaviside}\left (t -1\right ) {\mathrm e}^{2 t -2}}{2}+\frac {\operatorname {Heaviside}\left (t -1\right ) {\mathrm e}^{3 t -3}}{3} \]

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} e^{2 t} \left (-1+e^t\right ) & t\leq 1 \\ \frac {1}{6}-e^{2 t}+e^{3 t}-\frac {1}{2} e^{2 t-2}+\frac {1}{3} e^{3 t-3} & \text {True} \\ \end {array} \\ \end {array} \]

6.18 problem 68

6.18.1 Existence and uniqueness analysis

6.18.2 Maple step by step solution