Internal problem ID [12310]
Internal file name [OUTPUT/10963_Saturday_September_30_2023_08_26_38_PM_1620668/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 2(c).
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]]
\[ \boxed {4 y^{\prime \prime }+5 y^{\prime }+4 y=3 \,{\mathrm e}^{-t}} \] With initial conditions \begin {align*} [y \left (0\right ) = -1, 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) &={\frac {5}{4}}\\ q(t) &=1\\ F &=\frac {3 \,{\mathrm e}^{-t}}{4} \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+\frac {5 y^{\prime }}{4}+y = \frac {3 \,{\mathrm e}^{-t}}{4} \end {align*}
The domain of \(p(t)={\frac {5}{4}}\) 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*} 4 s^{2} Y \left (s \right )-4 y^{\prime }\left (0\right )-4 s y \left (0\right )+5 s Y \left (s \right )-5 y \left (0\right )+4 Y \left (s \right ) = \frac {3}{s +1}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=-1\\ y'(0) &=1 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} 4 s^{2} Y \left (s \right )+1+4 s +5 s Y \left (s \right )+4 Y \left (s \right ) = \frac {3}{s +1} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {4 s^{2}+5 s -2}{\left (s +1\right ) \left (4 s^{2}+5 s +4\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{s +1}+\frac {-1-\frac {i \sqrt {39}}{13}}{s +\frac {5}{8}-\frac {i \sqrt {39}}{8}}+\frac {-1+\frac {i \sqrt {39}}{13}}{s +\frac {5}{8}+\frac {i \sqrt {39}}{8}} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{s +1}\right ) &= {\mathrm e}^{-t}\\ \mathcal {L}^{-1}\left (\frac {-1-\frac {i \sqrt {39}}{13}}{s +\frac {5}{8}-\frac {i \sqrt {39}}{8}}\right ) &= -\frac {\left (i \sqrt {39}+13\right ) {\mathrm e}^{-\frac {\left (-i \sqrt {39}+5\right ) \left (-\frac {5 t}{8 \left (-\frac {5}{8}+\frac {i \sqrt {39}}{8}\right )}+\frac {i \sqrt {39}\, t}{-5+i \sqrt {39}}\right )}{8}}}{13}\\ \mathcal {L}^{-1}\left (\frac {-1+\frac {i \sqrt {39}}{13}}{s +\frac {5}{8}+\frac {i \sqrt {39}}{8}}\right ) &= -\frac {\left (13-i \sqrt {39}\right ) {\mathrm e}^{-\frac {\left (i \sqrt {39}+5\right ) \left (-\frac {5 t}{8 \left (-\frac {5}{8}-\frac {i \sqrt {39}}{8}\right )}-\frac {i \sqrt {39}\, t}{8 \left (-\frac {5}{8}-\frac {i \sqrt {39}}{8}\right )}\right )}{8}}}{13} \end {align*}
Adding the above results and simplifying gives \[ y={\mathrm e}^{-t}+\frac {2 \left (\sqrt {39}\, \sin \left (\frac {\sqrt {39}\, t}{8}\right )-13 \cos \left (\frac {\sqrt {39}\, t}{8}\right )\right ) {\mathrm e}^{-\frac {5 t}{8}}}{13} \] Simplifying the solution gives \[
y = \frac {2 \sin \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}} \sqrt {39}}{13}-2 \cos \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}+{\mathrm e}^{-t}
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {2 \sin \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}} \sqrt {39}}{13}-2 \cos \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}+{\mathrm e}^{-t} \\
\end{align*} Verification of solutions
\[
y = \frac {2 \sin \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}} \sqrt {39}}{13}-2 \cos \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}+{\mathrm e}^{-t}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [4 y^{\prime \prime }+5 y^{\prime }+4 y=3 \,{\mathrm e}^{-t}, y \left (0\right )=-1, 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 {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {5 y^{\prime }}{4}-y+\frac {3 \,{\mathrm e}^{-t}}{4} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {5 y^{\prime }}{4}+y=\frac {3 \,{\mathrm e}^{-t}}{4} \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+\frac {5}{4} r +1=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-\frac {5}{4}\right )\pm \left (\sqrt {-\frac {39}{16}}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-\frac {5}{8}-\frac {\mathrm {I} \sqrt {39}}{8}, -\frac {5}{8}+\frac {\mathrm {I} \sqrt {39}}{8}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}} \\ \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 {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}+c_{2} \sin \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}+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 )=\frac {3 \,{\mathrm e}^{-t}}{4}\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 {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}} & \sin \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}} \\ -\frac {\sin \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}} \sqrt {39}}{8}-\frac {5 \cos \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}}{8} & \frac {\sqrt {39}\, \cos \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}}{8}-\frac {5 \sin \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}}{8} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\frac {\sqrt {39}\, {\mathrm e}^{-\frac {5 t}{4}}}{8} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=\frac {2 \sqrt {39}\, {\mathrm e}^{-\frac {5 t}{8}} \left (-\cos \left (\frac {\sqrt {39}\, t}{8}\right ) \left (\int {\mathrm e}^{-\frac {3 t}{8}} \sin \left (\frac {\sqrt {39}\, t}{8}\right )d t \right )+\sin \left (\frac {\sqrt {39}\, t}{8}\right ) \left (\int {\mathrm e}^{-\frac {3 t}{8}} \cos \left (\frac {\sqrt {39}\, t}{8}\right )d t \right )\right )}{13} \\ {} & \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 {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}+c_{2} \sin \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}+{\mathrm e}^{-t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}+c_{2} \sin \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}+{\mathrm e}^{-t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-1 \\ {} & {} & -1=c_{1} +1 \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {c_{1} \sqrt {39}\, \sin \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}}{8}-\frac {5 c_{1} \cos \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}}{8}+\frac {c_{2} \sqrt {39}\, \cos \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}}{8}-\frac {5 c_{2} \sin \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}}{8}-{\mathrm e}^{-t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-1-\frac {5 c_{1}}{8}+\frac {c_{2} \sqrt {39}}{8} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-2, c_{2} =\frac {2 \sqrt {39}}{13}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {2 \sin \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}} \sqrt {39}}{13}-2 \cos \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}+{\mathrm e}^{-t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {2 \sin \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}} \sqrt {39}}{13}-2 \cos \left (\frac {\sqrt {39}\, t}{8}\right ) {\mathrm e}^{-\frac {5 t}{8}}+{\mathrm e}^{-t} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 5.547 (sec). Leaf size: 36
\[
y \left (t \right ) = \frac {2 \sqrt {39}\, {\mathrm e}^{-\frac {5 t}{8}} \sin \left (\frac {\sqrt {39}\, t}{8}\right )}{13}-2 \,{\mathrm e}^{-\frac {5 t}{8}} \cos \left (\frac {\sqrt {39}\, t}{8}\right )+{\mathrm e}^{-t}
\]
✓ Solution by Mathematica
Time used: 0.056 (sec). Leaf size: 58
\[
y(t)\to e^{-t}+2 \sqrt {\frac {3}{13}} e^{-5 t/8} \sin \left (\frac {\sqrt {39} t}{8}\right )-2 e^{-5 t/8} \cos \left (\frac {\sqrt {39} t}{8}\right )
\]
4.3.2 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([4*diff(y(t),t$2)+5*diff(y(t),t)+4*y(t)=3*exp(-t),y(0) = -1, D(y)(0) = 1],y(t), singsol=all)
DSolve[{4*y''[t]+5*y'[t]+4*y[t]==3*Exp[-t],{y[0]==-1,y'[0]==1}},y[t],t,IncludeSingularSolutions -> True]