Internal problem ID [6510]
Internal file name [OUTPUT/5758_Sunday_June_05_2022_03_53_01_PM_80445232/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 Problesm for review and discovery. Section
A, Drill exercises. Page 309
Problem number: 4(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, _missing_x]]
\[ \boxed {y^{\prime \prime }+3 y^{\prime }+3 y=2} \] 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 )+3 Y \left (s \right ) = \frac {2}{s}\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} +3 Y \left (s \right ) = \frac {2}{s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {c_{1} s^{2}+3 s c_{1} +c_{2} s +2}{s \left (s^{2}+3 s +3\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {2}{3 s}+\frac {\left (-c_{1} -\frac {2 c_{2}}{3}+\frac {2}{3}\right ) \left (\frac {i \sqrt {3}}{2}-\frac {3}{2}\right )-c_{1} -c_{2} +\frac {2}{3}}{s -\frac {i \sqrt {3}}{2}+\frac {3}{2}}+\frac {\left (-c_{1} -\frac {2 c_{2}}{3}+\frac {2}{3}\right ) \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )-c_{1} -c_{2} +\frac {2}{3}}{s +\frac {3}{2}+\frac {i \sqrt {3}}{2}} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {2}{3 s}\right ) &= {\frac {2}{3}}\\ \mathcal {L}^{-1}\left (\frac {\left (-c_{1} -\frac {2 c_{2}}{3}+\frac {2}{3}\right ) \left (\frac {i \sqrt {3}}{2}-\frac {3}{2}\right )-c_{1} -c_{2} +\frac {2}{3}}{s -\frac {i \sqrt {3}}{2}+\frac {3}{2}}\right ) &= \frac {{\mathrm e}^{-\frac {\left (3-i \sqrt {3}\right ) \left (\frac {i t}{i \sqrt {3}-3}-\frac {t \sqrt {3}}{2 \left (\frac {i \sqrt {3}}{2}-\frac {3}{2}\right )}\right ) \sqrt {3}}{2}} \left (-2-2 i c_{2} \sqrt {3}+2 i \sqrt {3}+3 \left (1-i \sqrt {3}\right ) c_{1} \right )}{6}\\ \mathcal {L}^{-1}\left (\frac {\left (-c_{1} -\frac {2 c_{2}}{3}+\frac {2}{3}\right ) \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )-c_{1} -c_{2} +\frac {2}{3}}{s +\frac {3}{2}+\frac {i \sqrt {3}}{2}}\right ) &= \frac {{\mathrm e}^{-\frac {\left (3+i \sqrt {3}\right ) \left (-\frac {t \sqrt {3}}{2 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}-\frac {i t}{2 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}\right ) \sqrt {3}}{2}} \left (-2+2 i c_{2} \sqrt {3}-2 i \sqrt {3}+3 c_{1} \left (1+i \sqrt {3}\right )\right )}{6} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {2}{3}+\frac {\left (\cos \left (\frac {\sqrt {3}\, t}{2}\right ) \left (-2+3 c_{1} \right )+\sin \left (\frac {\sqrt {3}\, t}{2}\right ) \sqrt {3}\, \left (3 c_{1} +2 c_{2} -2\right )\right ) {\mathrm e}^{-\frac {3 t}{2}}}{3} \] Simplifying the solution gives \[ y = \left (c_{1} +\frac {2 c_{2}}{3}-\frac {2}{3}\right ) \sqrt {3}\, {\mathrm e}^{-\frac {3 t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right )+\frac {2}{3}+\frac {{\mathrm e}^{-\frac {3 t}{2}} \left (-2+3 c_{1} \right ) \cos \left (\frac {\sqrt {3}\, t}{2}\right )}{3} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{1} +\frac {2 c_{2}}{3}-\frac {2}{3}\right ) \sqrt {3}\, {\mathrm e}^{-\frac {3 t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right )+\frac {2}{3}+\frac {{\mathrm e}^{-\frac {3 t}{2}} \left (-2+3 c_{1} \right ) \cos \left (\frac {\sqrt {3}\, t}{2}\right )}{3} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{1} +\frac {2 c_{2}}{3}-\frac {2}{3}\right ) \sqrt {3}\, {\mathrm e}^{-\frac {3 t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right )+\frac {2}{3}+\frac {{\mathrm e}^{-\frac {3 t}{2}} \left (-2+3 c_{1} \right ) \cos \left (\frac {\sqrt {3}\, t}{2}\right )}{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+3 y^{\prime }+3 y=2 \\ \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 +3=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-3\right )\pm \left (\sqrt {-3}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-\frac {3}{2}-\frac {\mathrm {I} \sqrt {3}}{2}, \frac {\mathrm {I} \sqrt {3}}{2}-\frac {3}{2}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-\frac {3 t}{2}} \cos \left (\frac {\sqrt {3}\, t}{2}\right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-\frac {3 t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\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={\mathrm e}^{-\frac {3 t}{2}} \cos \left (\frac {\sqrt {3}\, t}{2}\right ) c_{1} +{\mathrm e}^{-\frac {3 t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right ) 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 )=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}^{-\frac {3 t}{2}} \cos \left (\frac {\sqrt {3}\, t}{2}\right ) & {\mathrm e}^{-\frac {3 t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right ) \\ -\frac {3 \,{\mathrm e}^{-\frac {3 t}{2}} \cos \left (\frac {\sqrt {3}\, t}{2}\right )}{2}-\frac {\sqrt {3}\, {\mathrm e}^{-\frac {3 t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{2} & -\frac {3 \,{\mathrm e}^{-\frac {3 t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{2}+\frac {\sqrt {3}\, {\mathrm e}^{-\frac {3 t}{2}} \cos \left (\frac {\sqrt {3}\, t}{2}\right )}{2} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\frac {\sqrt {3}\, {\mathrm e}^{-3 t}}{2} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=\frac {4 \sqrt {3}\, {\mathrm e}^{-\frac {3 t}{2}} \left (-\cos \left (\frac {\sqrt {3}\, t}{2}\right ) \left (\int \sin \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {3 t}{2}}d t \right )+\sin \left (\frac {\sqrt {3}\, t}{2}\right ) \left (\int \cos \left (\frac {\sqrt {3}\, t}{2}\right ) {\mathrm e}^{\frac {3 t}{2}}d t \right )\right )}{3} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {2}{3} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-\frac {3 t}{2}} \cos \left (\frac {\sqrt {3}\, t}{2}\right ) c_{1} +{\mathrm e}^{-\frac {3 t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right ) c_{2} +\frac {2}{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: 1.844 (sec). Leaf size: 48
dsolve(diff(y(t),t$2)+3*diff(y(t),t)+3*y(t)=2,y(t), singsol=all)
\[ y \left (t \right ) = \frac {2}{3}+\frac {\left (\cos \left (\frac {\sqrt {3}\, t}{2}\right ) \left (-2+3 y \left (0\right )\right )+\sin \left (\frac {\sqrt {3}\, t}{2}\right ) \sqrt {3}\, \left (2 D\left (y \right )\left (0\right )+3 y \left (0\right )-2\right )\right ) {\mathrm e}^{-\frac {3 t}{2}}}{3} \]
✓ Solution by Mathematica
Time used: 0.02 (sec). Leaf size: 51
DSolve[y''[t]+3*y'[t]+3*y[t]==2,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to c_2 e^{-3 t/2} \cos \left (\frac {\sqrt {3} t}{2}\right )+c_1 e^{-3 t/2} \sin \left (\frac {\sqrt {3} t}{2}\right )+\frac {2}{3} \]