Internal problem ID [12788]
Internal file name [OUTPUT/11441_Saturday_November_04_2023_08_47_20_AM_58456927/index.tex]
Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 5. The Laplace Transform Method. Exercises 5.2, page 248
Problem number: 5.
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]]
\[ \boxed {y^{\prime \prime }+9 y=2 \sin \left (3 x \right )} \] 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 )+9 Y \left (s \right ) = \frac {6}{s^{2}+9}\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} +9 Y \left (s \right ) = \frac {6}{s^{2}+9} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {c_{1} s^{3}+c_{2} s^{2}+9 s c_{1} +9 c_{2} +6}{\left (s^{2}+9\right )^{2}} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= -\frac {1}{6 \left (s -3 i\right )^{2}}-\frac {1}{6 \left (s +3 i\right )^{2}}+\frac {3 i \left (-\frac {c_{2}}{18}-\frac {1}{54}\right )+\frac {c_{1}}{2}}{s -3 i}+\frac {-3 i \left (-\frac {c_{2}}{18}-\frac {1}{54}\right )+\frac {c_{1}}{2}}{s +3 i} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {1}{6 \left (s -3 i\right )^{2}}\right ) &= -\frac {x \,{\mathrm e}^{3 i x}}{6}\\ \mathcal {L}^{-1}\left (-\frac {1}{6 \left (s +3 i\right )^{2}}\right ) &= -\frac {x \,{\mathrm e}^{-3 i x}}{6}\\ \mathcal {L}^{-1}\left (\frac {3 i \left (-\frac {c_{2}}{18}-\frac {1}{54}\right )+\frac {c_{1}}{2}}{s -3 i}\right ) &= \frac {\left (-3 i c_{2} +9 c_{1} -i\right ) {\mathrm e}^{3 i x}}{18}\\ \mathcal {L}^{-1}\left (\frac {-3 i \left (-\frac {c_{2}}{18}-\frac {1}{54}\right )+\frac {c_{1}}{2}}{s +3 i}\right ) &= \frac {\left (3 i c_{2} +9 c_{1} +i\right ) {\mathrm e}^{-3 i x}}{18} \end {align*}
Adding the above results and simplifying gives \[ y=-\frac {\cos \left (3 x \right ) \left (-3 c_{1} +x \right )}{3}+\frac {\sin \left (3 x \right ) \left (1+3 c_{2} \right )}{9} \] Simplifying the solution gives \[ y = \frac {\left (-3 x +9 c_{1} \right ) \cos \left (3 x \right )}{9}+\frac {\sin \left (3 x \right ) \left (1+3 c_{2} \right )}{9} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-3 x +9 c_{1} \right ) \cos \left (3 x \right )}{9}+\frac {\sin \left (3 x \right ) \left (1+3 c_{2} \right )}{9} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-3 x +9 c_{1} \right ) \cos \left (3 x \right )}{9}+\frac {\sin \left (3 x \right ) \left (1+3 c_{2} \right )}{9} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+9 y=2 \sin \left (3 x \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}+9=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-36}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3 \,\mathrm {I}, 3 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )=\cos \left (3 x \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=\sin \left (3 x \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right )+y_{p}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y=\cos \left (3 x \right ) c_{1} +c_{2} \sin \left (3 x \right )+y_{p}\left (x \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (x \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 (x \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (x \right )=-y_{1}\left (x \right ) \left (\int \frac {y_{2}\left (x \right ) f \left (x \right )}{W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )}d x \right )+y_{2}\left (x \right ) \left (\int \frac {y_{1}\left (x \right ) f \left (x \right )}{W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )}d x \right ), f \left (x \right )=2 \sin \left (3 x \right )\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=\left [\begin {array}{cc} \cos \left (3 x \right ) & \sin \left (3 x \right ) \\ -3 \sin \left (3 x \right ) & 3 \cos \left (3 x \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=3 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & y_{p}\left (x \right )=-\frac {2 \cos \left (3 x \right ) \left (\int \sin \left (3 x \right )^{2}d x \right )}{3}+\frac {\sin \left (3 x \right ) \left (\int \sin \left (6 x \right )d x \right )}{3} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=\frac {\sin \left (3 x \right )}{18}-\frac {\cos \left (3 x \right ) x}{3} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=\cos \left (3 x \right ) c_{1} +c_{2} \sin \left (3 x \right )+\frac {\sin \left (3 x \right )}{18}-\frac {\cos \left (3 x \right ) x}{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: 5.437 (sec). Leaf size: 29
dsolve(diff(y(x),x$2)+9*y(x)=2*sin(3*x),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\cos \left (3 x \right ) \left (x -3 y \left (0\right )\right )}{3}+\frac {\sin \left (3 x \right ) \left (1+3 D\left (y \right )\left (0\right )\right )}{9} \]
✓ Solution by Mathematica
Time used: 0.051 (sec). Leaf size: 33
DSolve[y''[x]+9*y[x]==2*Sin[3*x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \left (-\frac {x}{3}+c_1\right ) \cos (3 x)+\frac {1}{18} (1+18 c_2) \sin (3 x) \]