Internal problem ID [13223]
Internal file name [OUTPUT/11879_Tuesday_December_05_2023_12_12_38_PM_64086885/index.tex]
Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th
edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 6. Laplace transform. Section 6.3 page 600
Problem number: 28.
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]]
\[ \boxed {y^{\prime \prime }-4 y={\mathrm e}^{2 t}} \] 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) &=0\\ q(t) &=-4\\ F &={\mathrm e}^{2 t} \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-4 y = {\mathrm e}^{2 t} \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 )-4 Y \left (s \right ) = \frac {1}{s -2}\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*} s^{2} Y \left (s \right )+1-s -4 Y \left (s \right ) = \frac {1}{s -2} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s^{2}-3 s +3}{\left (s -2\right ) \left (s^{2}-4\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {3}{16 \left (s -2\right )}+\frac {13}{16 \left (s +2\right )}+\frac {1}{4 \left (s -2\right )^{2}} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {3}{16 \left (s -2\right )}\right ) &= \frac {3 \,{\mathrm e}^{2 t}}{16}\\ \mathcal {L}^{-1}\left (\frac {13}{16 \left (s +2\right )}\right ) &= \frac {13 \,{\mathrm e}^{-2 t}}{16}\\ \mathcal {L}^{-1}\left (\frac {1}{4 \left (s -2\right )^{2}}\right ) &= \frac {t \,{\mathrm e}^{2 t}}{4} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {13 \,{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t} \left (3+4 t \right )}{16} \] Simplifying the solution gives \[
y = \frac {13 \,{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t} \left (3+4 t \right )}{16}
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {13 \,{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t} \left (3+4 t \right )}{16} \\
\end{align*} Verification of solutions
\[
y = \frac {13 \,{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t} \left (3+4 t \right )}{16}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }-4 y={\mathrm e}^{2 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 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}-4=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -2\right ) \left (r +2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2, 2\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-2 t} \\ \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} {\mathrm e}^{-2 t}+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 )={\mathrm e}^{2 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} {\mathrm e}^{-2 t} & {\mathrm e}^{2 t} \\ -2 \,{\mathrm e}^{-2 t} & 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 )=4 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {{\mathrm e}^{-2 t} \left (\int {\mathrm e}^{4 t}d t \right )}{4}+\frac {{\mathrm e}^{2 t} \left (\int 1d t \right )}{4} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {{\mathrm e}^{2 t} \left (-1+4 t \right )}{16} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{2 t}+\frac {{\mathrm e}^{2 t} \left (-1+4 t \right )}{16} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{2 t}+\frac {{\mathrm e}^{2 t} \left (-1+4 t \right )}{16} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} +c_{2} -\frac {1}{16} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} {\mathrm e}^{-2 t}+2 c_{2} {\mathrm e}^{2 t}+\frac {{\mathrm e}^{2 t} \left (-1+4 t \right )}{8}+\frac {{\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_{1} +2 c_{2} +\frac {1}{8} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {13}{16}, c_{2} =\frac {1}{4}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {13 \,{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t} \left (3+4 t \right )}{16} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {13 \,{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t} \left (3+4 t \right )}{16} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 5.0 (sec). Leaf size: 22
\[
y \left (t \right ) = \frac {13 \,{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t} \left (4 t +3\right )}{16}
\]
✓ Solution by Mathematica
Time used: 0.031 (sec). Leaf size: 27
\[
y(t)\to \frac {1}{16} e^{-2 t} \left (e^{4 t} (4 t+3)+13\right )
\]
19.2.2 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)-4*y(t)=exp(2*t),y(0) = 1, D(y)(0) = -1],y(t), singsol=all)
DSolve[{y''[t]-4*y[t]==Exp[2*t],{y[0]==1,y'[0]==-1}},y[t],t,IncludeSingularSolutions -> True]