Internal problem ID [13904]
Internal file name [OUTPUT/13076_Friday_February_23_2024_06_54_33_AM_94304757/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 31. Delta Functions. Additional Exercises. page 572
Problem number: 31.7 (g).
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]]
\[ \boxed {y^{\prime \prime }+4 y^{\prime }-12 y=\delta \left (t \right )} \] 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) &=4\\ q(t) &=-12\\ F &=\delta \left (t \right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+4 y^{\prime }-12 y = \delta \left (t \right ) \end {align*}
The domain of \(p(t)=4\) 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 )+4 s Y \left (s \right )-4 y \left (0\right )-12 Y \left (s \right ) = 1\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 )+4 s Y \left (s \right )-12 Y \left (s \right ) = 1 \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {1}{s^{2}+4 s -12} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{8 s -16}-\frac {1}{8 \left (s +6\right )} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{8 s -16}\right ) &= \frac {{\mathrm e}^{2 t}}{8}\\ \mathcal {L}^{-1}\left (-\frac {1}{8 \left (s +6\right )}\right ) &= -\frac {{\mathrm e}^{-6 t}}{8} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {{\mathrm e}^{-2 t} \sinh \left (4 t \right )}{4} \] Simplifying the solution gives \[
y = \frac {{\mathrm e}^{-2 t} \sinh \left (4 t \right )}{4}
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {{\mathrm e}^{-2 t} \sinh \left (4 t \right )}{4} \\
\end{align*} Verification of solutions
\[
y = \frac {{\mathrm e}^{-2 t} \sinh \left (4 t \right )}{4}
\] Warning, solution could not be verified \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+4 y^{\prime }-12 y=\mathit {Dirac}\left (t \right ), 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}+4 r -12=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +6\right ) \left (r -2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-6, 2\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}^{2 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}^{2 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 )=\mathit {Dirac}\left (t \right )\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}^{2 t} \\ -6 \,{\mathrm e}^{-6 t} & 2 \,{\mathrm e}^{2 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=8 \,{\mathrm e}^{-4 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=\frac {\left (\int \mathit {Dirac}\left (t \right )d t \right ) \left ({\mathrm e}^{8 t}-1\right ) {\mathrm e}^{-6 t}}{8} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {\mathit {Heaviside}\left (t \right ) \left ({\mathrm e}^{8 t}-1\right ) {\mathrm e}^{-6 t}}{8} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-6 t}+c_{2} {\mathrm e}^{2 t}+\frac {\mathit {Heaviside}\left (t \right ) \left ({\mathrm e}^{8 t}-1\right ) {\mathrm e}^{-6 t}}{8} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-6 t}+c_{2} {\mathrm e}^{2 t}+\frac {\mathit {Heaviside}\left (t \right ) \left ({\mathrm e}^{8 t}-1\right ) {\mathrm e}^{-6 t}}{8} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-6 c_{1} {\mathrm e}^{-6 t}+2 c_{2} {\mathrm e}^{2 t}+\frac {\mathit {Dirac}\left (t \right ) \left ({\mathrm e}^{8 t}-1\right ) {\mathrm e}^{-6 t}}{8}+\mathit {Heaviside}\left (t \right ) {\mathrm e}^{-6 t} {\mathrm e}^{8 t}-\frac {3 \mathit {Heaviside}\left (t \right ) \left ({\mathrm e}^{8 t}-1\right ) {\mathrm e}^{-6 t}}{4} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-6 c_{1} +2 c_{2} +1 \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {1}{8}, c_{2} =-\frac {1}{8}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left (\mathit {Heaviside}\left (t \right )-1\right ) \left ({\mathrm e}^{8 t}-1\right ) {\mathrm e}^{-6 t}}{8} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left (\mathit {Heaviside}\left (t \right )-1\right ) \left ({\mathrm e}^{8 t}-1\right ) {\mathrm e}^{-6 t}}{8} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 5.0 (sec). Leaf size: 14
\[
y \left (t \right ) = \frac {{\mathrm e}^{-2 t} \sinh \left (4 t \right )}{4}
\]
✓ Solution by Mathematica
Time used: 0.029 (sec). Leaf size: 28
\[
y(t)\to -\frac {1}{8} e^{-6 t} \left (e^{8 t}-1\right ) (\theta (0)-\theta (t))
\]
22.14.2 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)+4*diff(y(t),t)-12*y(t)=Dirac(t),y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
DSolve[{y''[t]+4*y'[t]-12*y[t]==DiracDelta[t],{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]