Internal problem ID [6501]
Internal file name [OUTPUT/5749_Sunday_June_05_2022_03_52_40_PM_91277379/index.tex
]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 7. Laplace Transforms. Section 7.5 The Unit Step and Impulse Functions. Page
303
Problem number: 1(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]]
\[ \boxed {y^{\prime \prime }-y=t^{2}} \] 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) &=-1\\ F &=t^{2} \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-y = t^{2} \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 )-Y \left (s \right ) = \frac {2}{s^{3}}\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 )-Y \left (s \right ) = \frac {2}{s^{3}} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {2}{s^{3} \left (s^{2}-1\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= -\frac {2}{s}+\frac {1}{s +1}-\frac {2}{s^{3}}+\frac {1}{s -1} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {2}{s}\right ) &= -2\\ \mathcal {L}^{-1}\left (\frac {1}{s +1}\right ) &= {\mathrm e}^{-t}\\ \mathcal {L}^{-1}\left (-\frac {2}{s^{3}}\right ) &= -t^{2}\\ \mathcal {L}^{-1}\left (\frac {1}{s -1}\right ) &= {\mathrm e}^{t} \end {align*}
Adding the above results and simplifying gives \[ y=-2-t^{2}+2 \cosh \left (t \right ) \] Simplifying the solution gives \[
y = -2-t^{2}+2 \cosh \left (t \right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= -2-t^{2}+2 \cosh \left (t \right ) \\
\end{align*} Verification of solutions
\[
y = -2-t^{2}+2 \cosh \left (t \right )
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-y=t^{2}, 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}-1=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -1\right ) \left (r +1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1, 1\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-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={\mathrm e}^{-t} c_{1} +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 )=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}^{-t} & {\mathrm e}^{t} \\ -{\mathrm e}^{-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 )=2 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {{\mathrm e}^{-t} \left (\int {\mathrm e}^{t} t^{2}d t \right )}{2}+\frac {{\mathrm e}^{t} \left (\int {\mathrm e}^{-t} t^{2}d t \right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-t^{2}-2 \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-t} c_{1} +c_{2} {\mathrm e}^{t}-t^{2}-2 \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y={\mathrm e}^{-t} c_{1} +c_{2} {\mathrm e}^{t}-t^{2}-2 \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} -2+c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-{\mathrm e}^{-t} c_{1} +c_{2} {\mathrm e}^{t}-2 t \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-c_{1} +c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =1, c_{2} =1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-t}+{\mathrm e}^{t}-t^{2}-2 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-t}+{\mathrm e}^{t}-t^{2}-2 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 1.516 (sec). Leaf size: 15
\[
y \left (t \right ) = -2-t^{2}+2 \cosh \left (t \right )
\]
✓ Solution by Mathematica
Time used: 0.015 (sec). Leaf size: 20
\[
y(t)\to -t^2+e^{-t}+e^t-2
\]
24.3.2 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)-y(t)=t^2,y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
DSolve[{y''[t]-y[t]==t^2,{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]