Internal problem ID [13899]
Internal file name [OUTPUT/13071_Friday_February_23_2024_06_54_31_AM_73885979/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 (b).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_laplace", "exact linear second order ode", "second_order_integrable_as_is", "second_order_ode_missing_y", "second_order_linear_constant_coeff"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {y^{\prime \prime }+3 y^{\prime }=\delta \left (t \right )} \] Since no initial conditions are explicitly given, then let \begin {align*} y \left (0\right )&= c_{1} \\ y'(0) &= c_{2} \end {align*}
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 )+3 s Y \left (s \right )-3 y \left (0\right ) = 1\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=c_{1}\\ y'(0) &=c_{2} \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-c_{2} -s c_{1} +3 s Y \left (s \right )-3 c_{1} = 1 \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s c_{1} +3 c_{1} +c_{2} +1}{s \left (s +3\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {c_{1} +\frac {c_{2}}{3}+\frac {1}{3}}{s}+\frac {-\frac {c_{2}}{3}-\frac {1}{3}}{s +3} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {c_{1} +\frac {c_{2}}{3}+\frac {1}{3}}{s}\right ) &= c_{1} +\frac {c_{2}}{3}+\frac {1}{3}\\ \mathcal {L}^{-1}\left (\frac {-\frac {c_{2}}{3}-\frac {1}{3}}{s +3}\right ) &= -\frac {\left (c_{2} +1\right ) {\mathrm e}^{-3 t}}{3} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {1}{3}+c_{1} +\frac {c_{2}}{3}-\frac {\left (c_{2} +1\right ) {\mathrm e}^{-3 t}}{3} \] Simplifying the solution gives \[ y = \frac {\left (-c_{2} -1\right ) {\mathrm e}^{-3 t}}{3}+c_{1} +\frac {c_{2}}{3}+\frac {1}{3} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-c_{2} -1\right ) {\mathrm e}^{-3 t}}{3}+c_{1} +\frac {c_{2}}{3}+\frac {1}{3} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-c_{2} -1\right ) {\mathrm e}^{-3 t}}{3}+c_{1} +\frac {c_{2}}{3}+\frac {1}{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+3 y^{\prime }=\mathit {Dirac}\left (t \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}+3 r =0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & r \left (r +3\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3, 0\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-3 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=1 \\ \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}^{-3 t}+c_{2} +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}^{-3 t} & 1 \\ -3 \,{\mathrm e}^{-3 t} & 0 \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=3 \,{\mathrm e}^{-3 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 (-1+{\mathrm e}^{-3 t}\right )}{3} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {\mathit {Heaviside}\left (t \right ) \left (-1+{\mathrm e}^{-3 t}\right )}{3} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-3 t}+c_{2} -\frac {\mathit {Heaviside}\left (t \right ) \left (-1+{\mathrm e}^{-3 t}\right )}{3} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable -> Calling odsolve with the ODE`, diff(_b(_a), _a) = -3*_b(_a)+Dirac(_a), _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful <- high order exact linear fully integrable successful`
✓ Solution by Maple
Time used: 5.031 (sec). Leaf size: 22
dsolve(diff(y(t),t$2)+3*diff(y(t),t)=Dirac(t),y(t), singsol=all)
\[ y \left (t \right ) = \frac {1}{3}+\frac {D\left (y \right )\left (0\right )}{3}+y \left (0\right )-\frac {{\mathrm e}^{-3 t} \left (1+D\left (y \right )\left (0\right )\right )}{3} \]
✓ Solution by Mathematica
Time used: 0.138 (sec). Leaf size: 27
DSolve[y''[t]+3*y'[t]==DiracDelta[t],y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {1}{3} \left (\theta (t)-e^{-3 t} (\theta (t)+c_1)\right )+c_2 \]