Internal problem ID [5696]
Internal file name [OUTPUT/4944_Sunday_June_05_2022_03_11_12_PM_3379136/index.tex]
Book: ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section: Chapter 6. Laplace Transforms. Problem set 6.3, page 224
Problem number: 20.
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]]
\[ \boxed {y^{\prime \prime }+10 y^{\prime }+24 y=144 t^{2}} \] With initial conditions \begin {align*} \left [y \left (0\right ) = {\frac {19}{12}}, y^{\prime }\left (0\right ) = -5\right ] \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) &=10\\ q(t) &=24\\ F &=144 t^{2} \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+10 y^{\prime }+24 y = 144 t^{2} \end {align*}
The domain of \(p(t)=10\) 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 )+10 s Y \left (s \right )-10 y \left (0\right )+24 Y \left (s \right ) = \frac {288}{s^{3}}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&={\frac {19}{12}}\\ y'(0) &=-5 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-\frac {65}{6}-\frac {19 s}{12}+10 s Y \left (s \right )+24 Y \left (s \right ) = \frac {288}{s^{3}} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {19 s^{2}-60 s +144}{12 s^{3}} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= -\frac {5}{s^{2}}+\frac {19}{12 s}+\frac {12}{s^{3}} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {5}{s^{2}}\right ) &= -5 t\\ \mathcal {L}^{-1}\left (\frac {19}{12 s}\right ) &= {\frac {19}{12}}\\ \mathcal {L}^{-1}\left (\frac {12}{s^{3}}\right ) &= 6 t^{2} \end {align*}
Adding the above results and simplifying gives \[ y=6 t^{2}-5 t +\frac {19}{12} \] Simplifying the solution gives \[
y = 6 t^{2}-5 t +\frac {19}{12}
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= 6 t^{2}-5 t +\frac {19}{12} \\
\end{align*} Verification of solutions
\[
y = 6 t^{2}-5 t +\frac {19}{12}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+10 y^{\prime }+24 y=144 t^{2}, y \left (0\right )=\frac {19}{12}, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-5\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}+10 r +24=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +6\right ) \left (r +4\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-6, -4\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-6 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-4 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}^{-6 t}+c_{2} {\mathrm e}^{-4 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 )=144 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}^{-6 t} & {\mathrm e}^{-4 t} \\ -6 \,{\mathrm e}^{-6 t} & -4 \,{\mathrm e}^{-4 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=2 \,{\mathrm e}^{-10 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-72 \,{\mathrm e}^{-6 t} \left (\int t^{2} {\mathrm e}^{6 t}d t \right )+72 \,{\mathrm e}^{-4 t} \left (\int t^{2} {\mathrm e}^{4 t}d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=6 t^{2}-5 t +\frac {19}{12} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-6 t}+c_{2} {\mathrm e}^{-4 t}+6 t^{2}-5 t +\frac {19}{12} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-6 t}+c_{2} {\mathrm e}^{-4 t}+6 t^{2}-5 t +\frac {19}{12} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=\frac {19}{12} \\ {} & {} & \frac {19}{12}=c_{1} +c_{2} +\frac {19}{12} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-6 c_{1} {\mathrm e}^{-6 t}-4 c_{2} {\mathrm e}^{-4 t}+12 t -5 \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-5 \\ {} & {} & -5=-6 c_{1} -4 c_{2} -5 \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =0\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=6 t^{2}-5 t +\frac {19}{12} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=6 t^{2}-5 t +\frac {19}{12} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.687 (sec). Leaf size: 14
\[
y \left (t \right ) = 6 t^{2}-5 t +\frac {19}{12}
\]
✓ Solution by Mathematica
Time used: 0.016 (sec). Leaf size: 17
\[
y(t)\to 6 t^2-5 t+\frac {19}{12}
\]
7.3.2 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([diff(y(t),t$2)+10*diff(y(t),t)+24*y(t)=144*t^2,y(0) = 19/12, D(y)(0) = -5],y(t), singsol=all)
DSolve[{y''[t]+10*y'[t]+24*y[t]==144*t^2,{y[0]==19/12,y'[0]==-5}},y[t],t,IncludeSingularSolutions -> True]