Internal problem ID [13869]
Internal file name [OUTPUT/13041_Friday_February_23_2024_06_54_19_AM_7363715/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 28. The inverse Laplace transform. Additional Exercises. page 509
Problem number: 28.9 (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_integrable_as_is", "second_order_ode_missing_y", "second_order_ode_quadrature", "second_order_linear_constant_coeff"
Maple gives the following as the ode type
[[_2nd_order, _quadrature]]
\[ \boxed {y^{\prime \prime }={\mathrm e}^{t} \sin \left (t \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0] \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) &=0\\ q(t) &=0\\ F &={\mathrm e}^{t} \sin \left (t \right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime } = {\mathrm e}^{t} \sin \left (t \right ) \end {align*}
The domain of \(p(t)=0\) 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 ) = \frac {1}{\left (s -1\right )^{2}+1}\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 ) = \frac {1}{\left (s -1\right )^{2}+1} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {1}{\left (s^{2}-2 s +2\right ) s^{2}} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= -\frac {1}{4 \left (s -1-i\right )}-\frac {1}{4 \left (s -1+i\right )}+\frac {1}{2 s}+\frac {1}{2 s^{2}} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {1}{4 \left (s -1-i\right )}\right ) &= -\frac {{\mathrm e}^{\left (1+i\right ) t}}{4}\\ \mathcal {L}^{-1}\left (-\frac {1}{4 \left (s -1+i\right )}\right ) &= -\frac {{\mathrm e}^{\left (1-i\right ) t}}{4}\\ \mathcal {L}^{-1}\left (\frac {1}{2 s}\right ) &= {\frac {1}{2}}\\ \mathcal {L}^{-1}\left (\frac {1}{2 s^{2}}\right ) &= \frac {t}{2} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {t}{2}+\frac {1}{2}-\frac {{\mathrm e}^{t} \cos \left (t \right )}{2} \] Simplifying the solution gives \[
y = \frac {t}{2}+\frac {1}{2}-\frac {{\mathrm e}^{t} \cos \left (t \right )}{2}
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {t}{2}+\frac {1}{2}-\frac {{\mathrm e}^{t} \cos \left (t \right )}{2} \\
\end{align*} Verification of solutions
\[
y = \frac {t}{2}+\frac {1}{2}-\frac {{\mathrm e}^{t} \cos \left (t \right )}{2}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }={\mathrm e}^{t} \sin \left (t \right ), 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 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {0}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=1 \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (t \right )\hspace {3pt}\textrm {by}\hspace {3pt} t \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=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} 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 )={\mathrm e}^{t} \sin \left (t \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 & t \\ 0 & 1 \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=1 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\left (\int t \,{\mathrm e}^{t} \sin \left (t \right )d t \right )+t \left (\int {\mathrm e}^{t} \sin \left (t \right )d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {{\mathrm e}^{t} \cos \left (t \right )}{2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} +c_{2} t -\frac {{\mathrm e}^{t} \cos \left (t \right )}{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} +c_{2} t -\frac {{\mathrm e}^{t} \cos \left (t \right )}{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} -\frac {1}{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{2} -\frac {{\mathrm e}^{t} \cos \left (t \right )}{2}+\frac {{\mathrm e}^{t} \sin \left (t \right )}{2} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=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 {1}{2}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {t}{2}+\frac {1}{2}-\frac {{\mathrm e}^{t} \cos \left (t \right )}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {t}{2}+\frac {1}{2}-\frac {{\mathrm e}^{t} \cos \left (t \right )}{2} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 4.703 (sec). Leaf size: 15
\[
y \left (t \right ) = -\frac {{\mathrm e}^{t} \cos \left (t \right )}{2}+\frac {t}{2}+\frac {1}{2}
\]
✓ Solution by Mathematica
Time used: 0.056 (sec). Leaf size: 19
\[
y(t)\to \frac {1}{2} \left (t-e^t \cos (t)+1\right )
\]
19.8.2 Maple step by step solution
`Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
<- quadrature successful`
dsolve([diff(y(t),t$2)=exp(t)*sin(t),y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
DSolve[{y''[t]==Exp[t]*Sin[t],{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]