Internal problem ID [13864]
Internal file name [OUTPUT/13036_Friday_February_23_2024_06_54_17_AM_48988675/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 28. The inverse Laplace transform. Additional Exercises. page 509
Problem number: 28.6 (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]]
\[ \boxed {y^{\prime \prime }+8 y^{\prime }+7 y=165 \,{\mathrm e}^{4 t}} \] With initial conditions \begin {align*} [y \left (0\right ) = 8, 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) &=8\\ q(t) &=7\\ F &=165 \,{\mathrm e}^{4 t} \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+8 y^{\prime }+7 y = 165 \,{\mathrm e}^{4 t} \end {align*}
The domain of \(p(t)=8\) is \[ \{-\infty <t <\infty \} \] And the point \(t_0 = 0\) is inside this domain. The domain of \(q(t)=7\) is \[ \{-\infty <t <\infty \} \] And the point \(t_0 = 0\) is also inside this domain. The domain of \(F =165 \,{\mathrm e}^{4 t}\) is \[ \{-\infty <t <\infty \} \] And the point \(t_0 = 0\) is also inside this domain. Hence solution exists and is unique.
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 )+8 s Y \left (s \right )-8 y \left (0\right )+7 Y \left (s \right ) = \frac {165}{s -4}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=8\\ y'(0) &=1 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-65-8 s +8 s Y \left (s \right )+7 Y \left (s \right ) = \frac {165}{s -4} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {8 s^{2}+33 s -95}{\left (s -4\right ) \left (s^{2}+8 s +7\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {3}{s -4}+\frac {1}{s +7}+\frac {4}{s +1} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {3}{s -4}\right ) &= 3 \,{\mathrm e}^{4 t}\\ \mathcal {L}^{-1}\left (\frac {1}{s +7}\right ) &= {\mathrm e}^{-7 t}\\ \mathcal {L}^{-1}\left (\frac {4}{s +1}\right ) &= 4 \,{\mathrm e}^{-t} \end {align*}
Adding the above results and simplifying gives \[ y={\mathrm e}^{-7 t}+4 \,{\mathrm e}^{-t}+3 \,{\mathrm e}^{4 t} \] Simplifying the solution gives \[ y = \left (3 \,{\mathrm e}^{11 t}+4 \,{\mathrm e}^{6 t}+1\right ) {\mathrm e}^{-7 t} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (3 \,{\mathrm e}^{11 t}+4 \,{\mathrm e}^{6 t}+1\right ) {\mathrm e}^{-7 t} \\ \end{align*}
Verification of solutions
\[ y = \left (3 \,{\mathrm e}^{11 t}+4 \,{\mathrm e}^{6 t}+1\right ) {\mathrm e}^{-7 t} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+8 y^{\prime }+7 y=165 \,{\mathrm e}^{4 t}, y \left (0\right )=8, 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}+8 r +7=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +7\right ) \left (r +1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-7, -1\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-7 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}^{-7 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 )=165 \,{\mathrm e}^{4 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}^{-7 t} & {\mathrm e}^{-t} \\ -7 \,{\mathrm e}^{-7 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 )=6 \,{\mathrm e}^{-8 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {55 \,{\mathrm e}^{-7 t} \left (\int {\mathrm e}^{11 t}d t \right )}{2}+\frac {55 \,{\mathrm e}^{-t} \left (\int {\mathrm e}^{5 t}d t \right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=3 \,{\mathrm e}^{4 t} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-7 t}+c_{2} {\mathrm e}^{-t}+3 \,{\mathrm e}^{4 t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-7 t}+c_{2} {\mathrm e}^{-t}+3 {\mathrm e}^{4 t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=8 \\ {} & {} & 8=c_{1} +c_{2} +3 \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-7 c_{1} {\mathrm e}^{-7 t}-c_{2} {\mathrm e}^{-t}+12 \,{\mathrm e}^{4 t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-7 c_{1} -c_{2} +12 \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =1, c_{2} =4\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\left (3 \,{\mathrm e}^{11 t}+4 \,{\mathrm e}^{6 t}+1\right ) {\mathrm e}^{-7 t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (3 \,{\mathrm e}^{11 t}+4 \,{\mathrm e}^{6 t}+1\right ) {\mathrm e}^{-7 t} \end {array} \]
Maple trace
`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`
✓ Solution by Maple
Time used: 4.906 (sec). Leaf size: 21
dsolve([diff(y(t),t$2)+8*diff(y(t),t)+7*y(t)=165*exp(4*t),y(0) = 8, D(y)(0) = 1],y(t), singsol=all)
\[ y \left (t \right ) = \left (3 \,{\mathrm e}^{11 t}+4 \,{\mathrm e}^{6 t}+1\right ) {\mathrm e}^{-7 t} \]
✓ Solution by Mathematica
Time used: 0.018 (sec). Leaf size: 25
DSolve[{y''[t]+8*y'[t]+7*y[t]==165*Exp[4*t],{y[0]==8,y'[0]==1}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to e^{-7 t}+4 e^{-t}+3 e^{4 t} \]