Internal problem ID [6508]
Internal file name [OUTPUT/5756_Sunday_June_05_2022_03_52_57_PM_36386801/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 Problesm for review and discovery. Section
A, Drill exercises. Page 309
Problem number: 3(d).
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]]
\[ \boxed {y^{\prime \prime }-y^{\prime }+y=3 \,{\mathrm e}^{-t}} \] With initial conditions \begin {align*} [y \left (0\right ) = 3, y^{\prime }\left (0\right ) = 2] \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) &=-1\\ q(t) &=1\\ F &=3 \,{\mathrm e}^{-t} \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-y^{\prime }+y = 3 \,{\mathrm e}^{-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 )+Y \left (s \right ) = \frac {3}{s +1}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=3\\ y'(0) &=2 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+1-3 s -s Y \left (s \right )+Y \left (s \right ) = \frac {3}{s +1} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {3 s^{2}+2 s +2}{\left (s +1\right ) \left (s^{2}-s +1\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {1-\frac {2 i \sqrt {3}}{3}}{s -\frac {1}{2}-\frac {i \sqrt {3}}{2}}+\frac {1+\frac {2 i \sqrt {3}}{3}}{s -\frac {1}{2}+\frac {i \sqrt {3}}{2}}+\frac {1}{s +1} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1-\frac {2 i \sqrt {3}}{3}}{s -\frac {1}{2}-\frac {i \sqrt {3}}{2}}\right ) &= \frac {\left (3-2 i \sqrt {3}\right ) {\mathrm e}^{\frac {\left (1+i \sqrt {3}\right ) t}{2}}}{3}\\ \mathcal {L}^{-1}\left (\frac {1+\frac {2 i \sqrt {3}}{3}}{s -\frac {1}{2}+\frac {i \sqrt {3}}{2}}\right ) &= \frac {\left (2 i \sqrt {3}+3\right ) {\mathrm e}^{\frac {\left (1-i \sqrt {3}\right ) t}{2}}}{3}\\ \mathcal {L}^{-1}\left (\frac {1}{s +1}\right ) &= {\mathrm e}^{-t} \end {align*}
Adding the above results and simplifying gives \[ y={\mathrm e}^{-t}+\frac {2 \,{\mathrm e}^{\frac {t}{2}} \left (2 \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, t}{2}\right )+3 \cos \left (\frac {\sqrt {3}\, t}{2}\right )\right )}{3} \] Simplifying the solution gives \[
y = \frac {\left (4 \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {3 t}{2}}+6 \cos \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {3 t}{2}}+3\right ) {\mathrm e}^{-t}}{3}
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {\left (4 \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {3 t}{2}}+6 \cos \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {3 t}{2}}+3\right ) {\mathrm e}^{-t}}{3} \\
\end{align*} Verification of solutions
\[
y = \frac {\left (4 \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {3 t}{2}}+6 \cos \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {3 t}{2}}+3\right ) {\mathrm e}^{-t}}{3}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-y^{\prime }+y=3 \,{\mathrm e}^{-t}, y \left (0\right )=3, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=2\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 +1=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {1\pm \left (\sqrt {-3}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}, \frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {t}{2}} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right ) \\ \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} \cos \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {t}{2}}+c_{2} {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right )+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 )=3 \,{\mathrm e}^{-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} \cos \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {t}{2}} & {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right ) \\ -\frac {\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {t}{2}}}{2}+\frac {\cos \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {t}{2}}}{2} & \frac {{\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{2}+\frac {{\mathrm e}^{\frac {t}{2}} \sqrt {3}\, \cos \left (\frac {\sqrt {3}\, t}{2}\right )}{2} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\frac {\sqrt {3}\, {\mathrm e}^{t}}{2} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=2 \sqrt {3}\, {\mathrm e}^{\frac {t}{2}} \left (-\cos \left (\frac {\sqrt {3}\, t}{2}\right ) \left (\int {\mathrm e}^{-\frac {3 t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right )d t \right )+\sin \left (\frac {\sqrt {3}\, t}{2}\right ) \left (\int {\mathrm e}^{-\frac {3 t}{2}} \cos \left (\frac {\sqrt {3}\, t}{2}\right )d t \right )\right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )={\mathrm e}^{-t} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {t}{2}}+c_{2} {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right )+{\mathrm e}^{-t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {t}{2}}+c_{2} {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right )+{\mathrm e}^{-t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 3=1+c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {c_{1} \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {t}{2}}}{2}+\frac {c_{1} \cos \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {t}{2}}}{2}+\frac {c_{2} {\mathrm e}^{\frac {t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{2}+\frac {c_{2} {\mathrm e}^{\frac {t}{2}} \sqrt {3}\, \cos \left (\frac {\sqrt {3}\, t}{2}\right )}{2}-{\mathrm e}^{-t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=2 \\ {} & {} & 2=-1+\frac {c_{1}}{2}+\frac {\sqrt {3}\, c_{2}}{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =2, c_{2} =\frac {4 \sqrt {3}}{3}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left (4 \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {3 t}{2}}+6 \cos \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {3 t}{2}}+3\right ) {\mathrm e}^{-t}}{3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left (4 \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {3 t}{2}}+6 \cos \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {3 t}{2}}+3\right ) {\mathrm e}^{-t}}{3} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 2.016 (sec). Leaf size: 37
\[
y \left (t \right ) = \frac {\left (4 \sqrt {3}\, {\mathrm e}^{\frac {3 t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right )+6 \,{\mathrm e}^{\frac {3 t}{2}} \cos \left (\frac {\sqrt {3}\, t}{2}\right )+3\right ) {\mathrm e}^{-t}}{3}
\]
✓ Solution by Mathematica
Time used: 0.03 (sec). Leaf size: 56
\[
y(t)\to e^{-t}+\frac {4 e^{t/2} \sin \left (\frac {\sqrt {3} t}{2}\right )}{\sqrt {3}}+2 e^{t/2} \cos \left (\frac {\sqrt {3} t}{2}\right )
\]
25.4.2 Maple step by step solution
`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`
dsolve([diff(y(t),t$2)-diff(y(t),t)+y(t)=3*exp(-t),y(0) = 3, D(y)(0) = 2],y(t), singsol=all)
DSolve[{y''[t]-y'[t]+y[t]==3*Exp[-t],{y[0]==3,y'[0]==2}},y[t],t,IncludeSingularSolutions -> True]