Internal problem ID [13863]
Internal file name [OUTPUT/13035_Friday_February_23_2024_06_54_17_AM_37563748/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 (b).
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, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime }+9 y=27 t^{3}} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0] \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) &=9\\ F &=27 t^{3} \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+9 y = 27 t^{3} \end {align*}
The domain of \(p(t)=0\) is \[ \{-\infty <t <\infty \} \] And the point \(t_0 = 0\) is inside this domain. The domain of \(q(t)=9\) is \[ \{-\infty <t <\infty \} \] And the point \(t_0 = 0\) is also inside this domain. The domain of \(F =27 t^{3}\) 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 )+9 Y \left (s \right ) = \frac {162}{s^{4}}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y'(0) &=0 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+9 Y \left (s \right ) = \frac {162}{s^{4}} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {162}{s^{4} \left (s^{2}+9\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= -\frac {i}{3 \left (s -3 i\right )}+\frac {i}{3 s +9 i}-\frac {2}{s^{2}}+\frac {18}{s^{4}} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {i}{3 \left (s -3 i\right )}\right ) &= -\frac {i {\mathrm e}^{3 i t}}{3}\\ \mathcal {L}^{-1}\left (\frac {i}{3 s +9 i}\right ) &= \frac {i {\mathrm e}^{-3 i t}}{3}\\ \mathcal {L}^{-1}\left (-\frac {2}{s^{2}}\right ) &= -2 t\\ \mathcal {L}^{-1}\left (\frac {18}{s^{4}}\right ) &= 3 t^{3} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {2 \sin \left (3 t \right )}{3}-2 t +3 t^{3} \] Simplifying the solution gives \[ y = \frac {2 \sin \left (3 t \right )}{3}-2 t +3 t^{3} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {2 \sin \left (3 t \right )}{3}-2 t +3 t^{3} \\ \end{align*}
Verification of solutions
\[ y = \frac {2 \sin \left (3 t \right )}{3}-2 t +3 t^{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+9 y=27 t^{3}, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0\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}+9=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-36}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3 \,\mathrm {I}, 3 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (3 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (3 t \right ) \\ \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 (3 t \right )+c_{2} \sin \left (3 t \right )+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 )=27 t^{3}\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 (3 t \right ) & \sin \left (3 t \right ) \\ -3 \sin \left (3 t \right ) & 3 \cos \left (3 t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=3 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-9 \cos \left (3 t \right ) \left (\int \sin \left (3 t \right ) t^{3}d t \right )+9 \sin \left (3 t \right ) \left (\int t^{3} \cos \left (3 t \right )d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=3 t^{3}-2 t \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (3 t \right )+c_{2} \sin \left (3 t \right )+3 t^{3}-2 t \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (3 t \right )+c_{2} \sin \left (3 t \right )+3 t^{3}-2 t \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 c_{1} \sin \left (3 t \right )+3 c_{2} \cos \left (3 t \right )+9 t^{2}-2 \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-2+3 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =\frac {2}{3}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {2 \sin \left (3 t \right )}{3}-2 t +3 t^{3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {2 \sin \left (3 t \right )}{3}-2 t +3 t^{3} \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: 4.156 (sec). Leaf size: 19
dsolve([diff(y(t),t$2)+9*y(t)=27*t^3,y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
\[ y \left (t \right ) = -2 t +\frac {2 \sin \left (3 t \right )}{3}+3 t^{3} \]
✓ Solution by Mathematica
Time used: 0.015 (sec). Leaf size: 22
DSolve[{y''[t]+9*y[t]==27*t^3,{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to 3 t^3-2 t+\frac {2}{3} \sin (3 t) \]