Internal problem ID [6704]
Internal file name [OUTPUT/5952_Sunday_June_05_2022_04_03_49_PM_95698925/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. EXERCISES 7.5. Page 315
Problem number: 8.
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_integrable_as_is", "second_order_ode_missing_y", "second_order_linear_constant_coeff"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {y^{\prime \prime }-2 y^{\prime }=1+\delta \left (t -2\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}
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 &=1+\delta \left (t -2\right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-2 y^{\prime } = 1+\delta \left (t -2\right ) \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 ) = \frac {1}{s}+{\mathrm e}^{-2 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 )-1-2 s Y \left (s \right ) = \frac {1}{s}+{\mathrm e}^{-2 s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {{\mathrm e}^{-2 s} s +s +1}{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 {{\mathrm e}^{-2 s} s +s +1}{s^{2} \left (s -2\right )}\right )\\ &= -\frac {3}{4}-\frac {\operatorname {Heaviside}\left (t -2\right )}{2}+\frac {3 \,{\mathrm e}^{2 t}}{4}-\frac {t}{2}+\frac {\left (1-\operatorname {Heaviside}\left (-t +2\right )\right ) {\mathrm e}^{2 t -4}}{2} \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} -\frac {3}{4}+\frac {3 \,{\mathrm e}^{2 t}}{4}-\frac {t}{2} & t <2 \\ -\frac {9}{4}+\frac {3 \,{\mathrm e}^{4}}{4} & t =2 \\ -\frac {5}{4}+\frac {3 \,{\mathrm e}^{2 t}}{4}-\frac {t}{2}+\frac {{\mathrm e}^{2 t -4}}{2} & 2 The solution(s) found are the following \begin{align*}
\tag{1} y &= -\frac {\left (\left \{\begin {array}{cc} 3-3 \,{\mathrm e}^{2 t}+2 t & t <2 \\ 9-3 \,{\mathrm e}^{4} & t &=2 \\ 5-3 \,{\mathrm e}^{2 t}+2 t -2 \,{\mathrm e}^{2 t -4} & 2 Verification of solutions
\[
y = -\frac {\left (\left \{\begin {array}{cc} 3-3 \,{\mathrm e}^{2 t}+2 t & t <2 \\ 9-3 \,{\mathrm e}^{4} & t =2 \\ 5-3 \,{\mathrm e}^{2 t}+2 t -2 \,{\mathrm e}^{2 t -4} & 2
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }-2 y^{\prime }=1+\mathit {Dirac}\left (t -2\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}-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 )=1+\mathit {Dirac}\left (t -2\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 )=-\frac {\left (\int \left (1+\mathit {Dirac}\left (t -2\right )\right )d t \right )}{2}+\frac {{\mathrm e}^{2 t} \left (\int \left ({\mathrm e}^{-4} \mathit {Dirac}\left (t -2\right )+{\mathrm e}^{-2 t}\right )d t \right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {t}{2}-\frac {\mathit {Heaviside}\left (t -2\right )}{2}+\frac {\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{2 t -4}}{2}-\frac {1}{4} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} +c_{2} {\mathrm e}^{2 t}-\frac {t}{2}-\frac {\mathit {Heaviside}\left (t -2\right )}{2}+\frac {\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{2 t -4}}{2}-\frac {1}{4} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} +c_{2} {\mathrm e}^{2 t}-\frac {t}{2}-\frac {\mathit {Heaviside}\left (t -2\right )}{2}+\frac {\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{2 t -4}}{2}-\frac {1}{4} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} -\frac {1}{4} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=2 c_{2} {\mathrm e}^{2 t}-\frac {1}{2}-\frac {\mathit {Dirac}\left (t -2\right )}{2}+\frac {\mathit {Dirac}\left (t -2\right ) {\mathrm e}^{2 t -4}}{2}+\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{2 t -4} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=2 c_{2} -\frac {1}{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {1}{2}, c_{2} =\frac {3}{4}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {3}{4}+\frac {3 \,{\mathrm e}^{2 t}}{4}-\frac {t}{2}-\frac {\mathit {Heaviside}\left (t -2\right )}{2}+\frac {\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{2 t -4}}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {3}{4}+\frac {3 \,{\mathrm e}^{2 t}}{4}-\frac {t}{2}-\frac {\mathit {Heaviside}\left (t -2\right )}{2}+\frac {\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{2 t -4}}{2} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 2.016 (sec). Leaf size: 39
\[
y \left (t \right ) = -\frac {3}{4}-\frac {\operatorname {Heaviside}\left (t -2\right )}{2}+\frac {3 \,{\mathrm e}^{2 t}}{4}-\frac {t}{2}+\frac {\operatorname {Heaviside}\left (t -2\right ) {\mathrm e}^{2 t -4}}{2}
\]
✓ Solution by Mathematica
Time used: 0.116 (sec). Leaf size: 37
\[
y(t)\to \frac {1}{4} \left (\left (2 e^{2 t-4}-2\right ) \theta (t-2)-2 t+3 e^{2 t}-3\right )
\]
8.8.2 Maple step by step solution
`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)+1+Dirac(_a-2), _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`
dsolve([diff(y(t),t$2)-2*diff(y(t),t)=1+Dirac(t-2),y(0) = 0, D(y)(0) = 1],y(t), singsol=all)
DSolve[{y''[t]-2*y'[t]==1+DiracDelta[t-2],{y[0]==0,y'[0]==1}},y[t],t,IncludeSingularSolutions -> True]