Internal problem ID [7310]
Internal file name [OUTPUT/6296_Sunday_June_05_2022_04_39_04_PM_47141029/index.tex]
Book: Own collection of miscellaneous problems
Section: section 5.0
Problem number: 17.
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]]
\[ \boxed {y^{\prime \prime }+3 y^{\prime }-4 y=6 \,{\mathrm e}^{2 t -2}} \] With initial conditions \begin {align*} [y \left (1\right ) = 4, y^{\prime }\left (1\right ) = 5] \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) &=-4\\ F &=6 \,{\mathrm e}^{2 t -2} \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+3 y^{\prime }-4 y = 6 \,{\mathrm e}^{2 t -2} \end {align*}
The domain of \(p(t)=3\) is \[
\{-\infty Since both initial conditions are not at zero, then let \begin {align*} y(0) &= c_{1}\\ y'(0) &= c_{2} \end {align*}
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 )-4 Y \left (s \right ) = \frac {6 \,{\mathrm e}^{-2}}{s -2}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=c_{1}\\ y'(0) &=c_{2} \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-c_{2} -s c_{1} +3 s Y \left (s \right )-3 c_{1} -4 Y \left (s \right ) = \frac {6 \,{\mathrm e}^{-2}}{s -2} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s^{2} c_{1} +s c_{1} +c_{2} s +6 \,{\mathrm e}^{-2}-6 c_{1} -2 c_{2}}{\left (s -2\right ) \left (s^{2}+3 s -4\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {{\mathrm e}^{-2}}{s -2}+\frac {\frac {c_{1}}{5}-\frac {c_{2}}{5}+\frac {{\mathrm e}^{-2}}{5}}{s +4}+\frac {\frac {4 c_{1}}{5}+\frac {c_{2}}{5}-\frac {6 \,{\mathrm e}^{-2}}{5}}{s -1} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {{\mathrm e}^{-2}}{s -2}\right ) &= {\mathrm e}^{2 t -2}\\ \mathcal {L}^{-1}\left (\frac {\frac {c_{1}}{5}-\frac {c_{2}}{5}+\frac {{\mathrm e}^{-2}}{5}}{s +4}\right ) &= \frac {\left (c_{1} -c_{2} +{\mathrm e}^{-2}\right ) {\mathrm e}^{-4 t}}{5}\\ \mathcal {L}^{-1}\left (\frac {\frac {4 c_{1}}{5}+\frac {c_{2}}{5}-\frac {6 \,{\mathrm e}^{-2}}{5}}{s -1}\right ) &= \frac {{\mathrm e}^{t} \left (4 c_{1} +c_{2} -6 \,{\mathrm e}^{-2}\right )}{5} \end {align*}
Adding the above results and simplifying gives \[ y={\mathrm e}^{2 t -2}+\frac {{\mathrm e}^{t} \left (4 c_{1} +c_{2} -6 \,{\mathrm e}^{-2}\right )}{5}+\frac {\left (c_{1} -c_{2} +{\mathrm e}^{-2}\right ) {\mathrm e}^{-4 t}}{5} \] Since both initial conditions given are not at
zero, then we need to setup two equations to solve for \(c_{1},c_{1}\). At \(t=1\) the first equation becomes, using
the above solution \begin {align*} 4 &= 1+\frac {{\mathrm e} \left (4 c_{1} +c_{2} -6 \,{\mathrm e}^{-2}\right )}{5}+\frac {\left (c_{1} -c_{2} +{\mathrm e}^{-2}\right ) {\mathrm e}^{-4}}{5} \end {align*}
And taking derivative of the solution and evaluating at \(t=1\) gives the second equation as
\begin {align*} 5 &= 2+\frac {{\mathrm e} \left (4 c_{1} +c_{2} -6 \,{\mathrm e}^{-2}\right )}{5}-\frac {4 \left (c_{1} -c_{2} +{\mathrm e}^{-2}\right ) {\mathrm e}^{-4}}{5} \end {align*}
Solving gives \begin {align*} c_{1} &= \left ({\mathrm e} \,{\mathrm e}^{-2}+3\right ) {\mathrm e}^{-1}\\ c_{2} &= {\mathrm e}^{-1} \left (2 \,{\mathrm e} \,{\mathrm e}^{-2}+3\right ) \end {align*}
Subtituting these in the solution obtained above gives \begin {align*} y &= {\mathrm e}^{2 t -2}+\frac {{\mathrm e}^{t} \left (4 \left ({\mathrm e} \,{\mathrm e}^{-2}+3\right ) {\mathrm e}^{-1}+{\mathrm e}^{-1} \left (2 \,{\mathrm e} \,{\mathrm e}^{-2}+3\right )-6 \,{\mathrm e}^{-2}\right )}{5}+\frac {\left (\left ({\mathrm e} \,{\mathrm e}^{-2}+3\right ) {\mathrm e}^{-1}-{\mathrm e}^{-1} \left (2 \,{\mathrm e} \,{\mathrm e}^{-2}+3\right )+{\mathrm e}^{-2}\right ) {\mathrm e}^{-4 t}}{5}\\ &= {\mathrm e}^{2 t -2}+3 \,{\mathrm e}^{t -1} \end {align*}
Simplifying the solution gives \[
y = {\mathrm e}^{2 t -2}+3 \,{\mathrm e}^{t -1}
\] The solution(s) found are the following \begin{align*}
\tag{1} y &= {\mathrm e}^{2 t -2}+3 \,{\mathrm e}^{t -1} \\
\end{align*} Verification of solutions
\[
y = {\mathrm e}^{2 t -2}+3 \,{\mathrm e}^{t -1}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+3 y^{\prime }-4 y=6 \,{\mathrm e}^{2 t -2}, y \left (1\right )=4, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}1\right \}}}}=5\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 -4=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +4\right ) \left (r -1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-4, 1\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-4 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{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}^{-4 t}+c_{2} {\mathrm e}^{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 )=6 \,{\mathrm e}^{2 t -2}\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}^{-4 t} & {\mathrm e}^{t} \\ -4 \,{\mathrm e}^{-4 t} & {\mathrm e}^{t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=5 \,{\mathrm e}^{-3 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=\frac {6 \left ({\mathrm e}^{5 t} \left (\int {\mathrm e}^{t -2}d t \right )-\left (\int {\mathrm e}^{6 t -2}d t \right )\right ) {\mathrm e}^{-4 t}}{5} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )={\mathrm e}^{2 t -2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-4 t}+c_{2} {\mathrm e}^{t}+{\mathrm e}^{2 t -2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-4 t}+c_{2} {\mathrm e}^{t}+{\mathrm e}^{2 t -2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=4 \\ {} & {} & 4=c_{1} {\mathrm e}^{-4}+c_{2} {\mathrm e}+1 \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-4 c_{1} {\mathrm e}^{-4 t}+c_{2} {\mathrm e}^{t}+2 \,{\mathrm e}^{2 t -2} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}1\right \}}}}=5 \\ {} & {} & 5=-4 c_{1} {\mathrm e}^{-4}+c_{2} {\mathrm e}+2 \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =\frac {3}{{\mathrm e}}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{2 t -2}+3 \,{\mathrm e}^{t -1} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{2 t -2}+3 \,{\mathrm e}^{t -1} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.297 (sec). Leaf size: 17
\[
y \left (t \right ) = {\mathrm e}^{2 t -2}+3 \,{\mathrm e}^{t -1}
\]
✓ Solution by Mathematica
Time used: 0.078 (sec). Leaf size: 18
\[
y(t)\to e^{t-2} \left (e^t+3 e\right )
\]
5.17.2 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)-4*y(t)=6*exp(2*t-2),y(1) = 4, D(y)(1) = 5],y(t), singsol=all)
DSolve[{y''[t]+3*y'[t]-4*y[t]==6*Exp[2*t-2],{y[1]==4,y'[1]==5}},y[t],t,IncludeSingularSolutions -> True]