problem Problem 4(a)

Internal problem ID [12332]
Internal ﬁle name [OUTPUT/10985_Monday_October_02_2023_02_47_43_AM_94070824/index.tex]

Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section: Chapter 5.6 Laplace transform. Nonhomogeneous equations. Problems page 368
Problem number: Problem 4(a).
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_laplace", "exact linear second order ode", "second_order_ode_missing_y", "second_order_linear_constant_coeﬀ"

\[ \boxed {y^{\prime \prime }-2 y^{\prime }=\left \{\begin {array}{cc} 4 & 0\le t <1 \\ 6 & 1\le t \end {array}\right .} \] With initial conditions \begin {align*} [y \left (0\right ) = -6, y^{\prime }\left (0\right ) = 1] \end {align*}

4.25.1 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) &=-2\\ q(t) &=0\\ F &=\left \{\begin {array}{cc} 0 & t <0 \\ 4 & t <1 \\ 6 & 1\le t \end {array}\right . \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-2 y^{\prime } = \left \{\begin {array}{cc} 0 & t <0 \\ 4 & t <1 \\ 6 & 1\le t \end {array}\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 )-2 s Y \left (s \right )+2 y \left (0\right ) = \frac {4+2 \,{\mathrm e}^{-s}}{s}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=-6\\ y'(0) &=1 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-13+6 s -2 s Y \left (s \right ) = \frac {4+2 \,{\mathrm e}^{-s}}{s} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {-6 s^{2}+2 \,{\mathrm e}^{-s}+13 s +4}{s^{2} \left (s -2\right )} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {-6 s^{2}+2 \,{\mathrm e}^{-s}+13 s +4}{s^{2} \left (s -2\right )}\right )\\ &= -\frac {15}{2}-2 t +\frac {3 \,{\mathrm e}^{2 t}}{2}+\frac {\left (1-\operatorname {Heaviside}\left (1-t \right )\right ) {\mathrm e}^{2 t -2}}{2}-\frac {\operatorname {Heaviside}\left (t -1\right ) \left (2 t -1\right )}{2} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} -\frac {15}{2}-2 t +\frac {3 \,{\mathrm e}^{2 t}}{2} & t <1 \\ -10+\frac {3 \,{\mathrm e}^{2}}{2} & t =1 \\ -7-3 t +\frac {3 \,{\mathrm e}^{2 t}}{2}+\frac {{\mathrm e}^{2 t -2}}{2} & 1

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (\left \{\begin {array}{cc} 15+4 t -3 \,{\mathrm e}^{2 t} & t <1 \\ 20-3 \,{\mathrm e}^{2} & t &=1 \\ 14+6 t -3 \,{\mathrm e}^{2 t}-{\mathrm e}^{2 t -2} & 1

Veriﬁcation of solutions

\[ y = -\frac {\left (\left \{\begin {array}{cc} 15+4 t -3 \,{\mathrm e}^{2 t} & t <1 \\ 20-3 \,{\mathrm e}^{2} & t =1 \\ 14+6 t -3 \,{\mathrm e}^{2 t}-{\mathrm e}^{2 t -2} & 1

4.25.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-2 y^{\prime }=\left \{\begin {array}{cc} 0 & t <0 \\ 4 & t <1 \\ 6 & 1\le t \end {array}\right ., y \left (0\right )=-6, 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}-2 r =0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & r \left (r -2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (0, 2\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=1 \\ \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} +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 )=\left \{\begin {array}{cc} 0 & t <0 \\ 4 & t <1 \\ 6 & 1\le t \end {array}\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} 1 & {\mathrm e}^{2 t} \\ 0 & 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 )=2 \,{\mathrm e}^{2 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\left (\int \left (\left \{\begin {array}{cc} 0 & t <0 \\ 2 & t <1 \\ 3 & 1\le t \end {array}\right .\right )d t \right )+\frac {{\mathrm e}^{2 t} \left (\int \left (\left \{\begin {array}{cc} 0 & t <0 \\ 4 & t <1 \\ 6 & 1\le t \end {array}\right .\right ) {\mathrm e}^{-2 t}d t \right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\left \{\begin {array}{cc} 0 & t \le 0 \\ -1+{\mathrm e}^{2 t}-2 t & t \le 1 \\ -3 t -\frac {1}{2}+\frac {{\mathrm e}^{2 t -2}}{2}+{\mathrm e}^{2 t} & 1

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
-> Calling odsolve with the ODE`, diff(_b(_a), _a) = 2*_b(_a)+4*Heaviside(_a)+2*Heaviside(_a-1), _b(_a)`   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful 
<- high order exact linear fully integrable successful`

\[ y \left (t \right ) = -\frac {\left (\left \{\begin {array}{cc} 15+4 t -3 \,{\mathrm e}^{2 t} & t <1 \\ 20-3 \,{\mathrm e}^{2} & t =1 \\ 14+6 t -3 \,{\mathrm e}^{2 t}-{\mathrm e}^{2 t -2} & 1

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {1}{2} \left (-13+e^{2 t}\right ) & t\leq 0 \\ \frac {1}{2} \left (-4 t+3 e^{2 t}-15\right ) & 0

4.25 problem Problem 4(a)

4.25.1 Existence and uniqueness analysis

4.25.2 Maple step by step solution