Internal problem ID [6505]
Internal file name [OUTPUT/5753_Sunday_June_05_2022_03_52_50_PM_67993605/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(a).
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, _missing_x]]
\[ \boxed {y^{\prime \prime }+3 y^{\prime }-5 y=1} \] 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) &=3\\ q(t) &=-5\\ F &=1 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+3 y^{\prime }-5 y = 1 \end {align*}
The domain of \(p(t)=3\) 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 )+3 s Y \left (s \right )-3 y \left (0\right )-5 Y \left (s \right ) = \frac {1}{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+3 s Y \left (s \right )-5 Y \left (s \right ) = \frac {1}{s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {1+s}{s \left (s^{2}+3 s -5\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {\frac {1}{10}+\frac {13 \sqrt {29}}{290}}{s +\frac {3}{2}-\frac {\sqrt {29}}{2}}+\frac {\frac {1}{10}-\frac {13 \sqrt {29}}{290}}{s +\frac {3}{2}+\frac {\sqrt {29}}{2}}-\frac {1}{5 s} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {\frac {1}{10}+\frac {13 \sqrt {29}}{290}}{s +\frac {3}{2}-\frac {\sqrt {29}}{2}}\right ) &= \frac {\left (29+13 \sqrt {29}\right ) {\mathrm e}^{-\frac {3 t}{2}+\frac {t \sqrt {29}}{2}}}{290}\\ \mathcal {L}^{-1}\left (\frac {\frac {1}{10}-\frac {13 \sqrt {29}}{290}}{s +\frac {3}{2}+\frac {\sqrt {29}}{2}}\right ) &= \frac {{\mathrm e}^{\frac {\left (-3-\sqrt {29}\right ) \left (-\frac {3 \left (3+\sqrt {29}\right ) t}{2 \left (-\frac {3}{2}-\frac {\sqrt {29}}{2}\right )}-\frac {\left (3+\sqrt {29}\right ) t \sqrt {29}}{2 \left (-\frac {3}{2}-\frac {\sqrt {29}}{2}\right )}\right )}{6+2 \sqrt {29}}} \left (29-13 \sqrt {29}\right )}{290}\\ \mathcal {L}^{-1}\left (-\frac {1}{5 s}\right ) &= -{\frac {1}{5}} \end {align*}
Adding the above results and simplifying gives \[ y=-\frac {1}{5}+\frac {{\mathrm e}^{-\frac {3 t}{2}} \left (13 \sqrt {29}\, \sinh \left (\frac {t \sqrt {29}}{2}\right )+29 \cosh \left (\frac {t \sqrt {29}}{2}\right )\right )}{145} \] Simplifying the solution gives \[
y = \frac {13 \sqrt {29}\, \sinh \left (\frac {t \sqrt {29}}{2}\right ) {\mathrm e}^{-\frac {3 t}{2}}}{145}+\frac {\cosh \left (\frac {t \sqrt {29}}{2}\right ) {\mathrm e}^{-\frac {3 t}{2}}}{5}-\frac {1}{5}
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {13 \sqrt {29}\, \sinh \left (\frac {t \sqrt {29}}{2}\right ) {\mathrm e}^{-\frac {3 t}{2}}}{145}+\frac {\cosh \left (\frac {t \sqrt {29}}{2}\right ) {\mathrm e}^{-\frac {3 t}{2}}}{5}-\frac {1}{5} \\
\end{align*} Verification of solutions
\[
y = \frac {13 \sqrt {29}\, \sinh \left (\frac {t \sqrt {29}}{2}\right ) {\mathrm e}^{-\frac {3 t}{2}}}{145}+\frac {\cosh \left (\frac {t \sqrt {29}}{2}\right ) {\mathrm e}^{-\frac {3 t}{2}}}{5}-\frac {1}{5}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+3 y^{\prime }-5 y=1, 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 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+3 r -5=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-3\right )\pm \left (\sqrt {29}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-\frac {3}{2}-\frac {\sqrt {29}}{2}, -\frac {3}{2}+\frac {\sqrt {29}}{2}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{\left (-\frac {3}{2}-\frac {\sqrt {29}}{2}\right ) t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{\left (-\frac {3}{2}+\frac {\sqrt {29}}{2}\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} {\mathrm e}^{\left (-\frac {3}{2}-\frac {\sqrt {29}}{2}\right ) t}+c_{2} {\mathrm e}^{\left (-\frac {3}{2}+\frac {\sqrt {29}}{2}\right ) 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\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}^{\left (-\frac {3}{2}-\frac {\sqrt {29}}{2}\right ) t} & {\mathrm e}^{\left (-\frac {3}{2}+\frac {\sqrt {29}}{2}\right ) t} \\ \left (-\frac {3}{2}-\frac {\sqrt {29}}{2}\right ) {\mathrm e}^{\left (-\frac {3}{2}-\frac {\sqrt {29}}{2}\right ) t} & \left (-\frac {3}{2}+\frac {\sqrt {29}}{2}\right ) {\mathrm e}^{\left (-\frac {3}{2}+\frac {\sqrt {29}}{2}\right ) t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\sqrt {29}\, {\mathrm e}^{-3 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {\sqrt {29}\, \left ({\mathrm e}^{-\frac {\left (3+\sqrt {29}\right ) t}{2}} \left (\int {\mathrm e}^{\frac {\left (3+\sqrt {29}\right ) t}{2}}d t \right )-{\mathrm e}^{\frac {\left (-3+\sqrt {29}\right ) t}{2}} \left (\int {\mathrm e}^{-\frac {\left (-3+\sqrt {29}\right ) t}{2}}d t \right )\right )}{29} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {1}{5} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{\left (-\frac {3}{2}-\frac {\sqrt {29}}{2}\right ) t}+c_{2} {\mathrm e}^{\left (-\frac {3}{2}+\frac {\sqrt {29}}{2}\right ) t}-\frac {1}{5} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{\left (-\frac {3}{2}-\frac {\sqrt {29}}{2}\right ) t}+c_{2} {\mathrm e}^{\left (-\frac {3}{2}+\frac {\sqrt {29}}{2}\right ) t}-\frac {1}{5} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} -\frac {1}{5} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{1} \left (-\frac {3}{2}-\frac {\sqrt {29}}{2}\right ) {\mathrm e}^{\left (-\frac {3}{2}-\frac {\sqrt {29}}{2}\right ) t}+c_{2} \left (-\frac {3}{2}+\frac {\sqrt {29}}{2}\right ) {\mathrm e}^{\left (-\frac {3}{2}+\frac {\sqrt {29}}{2}\right ) t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=c_{1} \left (-\frac {3}{2}-\frac {\sqrt {29}}{2}\right )+c_{2} \left (-\frac {3}{2}+\frac {\sqrt {29}}{2}\right ) \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {1}{10}-\frac {13 \sqrt {29}}{290}, c_{2} =\frac {1}{10}+\frac {13 \sqrt {29}}{290}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left (29+13 \sqrt {29}\right ) {\mathrm e}^{\frac {\left (-3+\sqrt {29}\right ) t}{2}}}{290}-\frac {1}{5}+\frac {\left (29-13 \sqrt {29}\right ) {\mathrm e}^{-\frac {\left (3+\sqrt {29}\right ) t}{2}}}{290} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left (29+13 \sqrt {29}\right ) {\mathrm e}^{\frac {\left (-3+\sqrt {29}\right ) t}{2}}}{290}-\frac {1}{5}+\frac {\left (29-13 \sqrt {29}\right ) {\mathrm e}^{-\frac {\left (3+\sqrt {29}\right ) t}{2}}}{290} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 1.953 (sec). Leaf size: 34
\[
y \left (t \right ) = \frac {13 \,{\mathrm e}^{-\frac {3 t}{2}} \sqrt {29}\, \sinh \left (\frac {t \sqrt {29}}{2}\right )}{145}+\frac {{\mathrm e}^{-\frac {3 t}{2}} \cosh \left (\frac {t \sqrt {29}}{2}\right )}{5}-\frac {1}{5}
\]
✓ Solution by Mathematica
Time used: 0.026 (sec). Leaf size: 67
\[
y(t)\to \frac {1}{290} e^{-\frac {1}{2} \left (3+\sqrt {29}\right ) t} \left (\left (29+13 \sqrt {29}\right ) e^{\sqrt {29} t}-58 e^{\frac {1}{2} \left (3+\sqrt {29}\right ) t}+29-13 \sqrt {29}\right )
\]
25.1.2 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)+3*diff(y(t),t)-5*y(t)=1,y(0) = 0, D(y)(0) = 1],y(t), singsol=all)
DSolve[{y''[t]+3*y'[t]-5*y[t]==1,{y[0]==0,y'[0]==1}},y[t],t,IncludeSingularSolutions -> True]