Internal problem ID [12800]
Internal file name [OUTPUT/11453_Saturday_November_04_2023_08_47_23_AM_38420457/index.tex]
Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 5. The Laplace Transform Method. Exercises 5.3, page 255
Problem number: 9.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeff", "second_order_ode_can_be_made_integrable"
Maple gives the following as the ode type
[[_2nd_order, _missing_x]]
\[ \boxed {y^{\prime \prime }+9 y=1} \] 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(x)y^{\prime } + q(x) y &= F \end {align*}
Where here \begin {align*} p(x) &=0\\ q(x) &=9\\ F &=1 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+9 y = 1 \end {align*}
The domain of \(p(x)=0\) 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 )+9 Y \left (s \right ) = \frac {1}{s}\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 )+9 Y \left (s \right ) = \frac {1}{s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {1}{s \left (s^{2}+9\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= -\frac {1}{18 \left (s -3 i\right )}-\frac {1}{18 \left (s +3 i\right )}+\frac {1}{9 s} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {1}{18 \left (s -3 i\right )}\right ) &= -\frac {{\mathrm e}^{3 i x}}{18}\\ \mathcal {L}^{-1}\left (-\frac {1}{18 \left (s +3 i\right )}\right ) &= -\frac {{\mathrm e}^{-3 i x}}{18}\\ \mathcal {L}^{-1}\left (\frac {1}{9 s}\right ) &= {\frac {1}{9}} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {1}{9}-\frac {\cos \left (3 x \right )}{9} \] Simplifying the solution gives \[
y = \frac {1}{9}-\frac {\cos \left (3 x \right )}{9}
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {1}{9}-\frac {\cos \left (3 x \right )}{9} \\
\end{align*} Verification of solutions
\[
y = \frac {1}{9}-\frac {\cos \left (3 x \right )}{9}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+9 y=1, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \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}+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 )=1\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 {\cos \left (3 x \right ) \left (\int \sin \left (3 x \right )d x \right )}{3}+\frac {\sin \left (3 x \right ) \left (\int \cos \left (3 x \right )d x \right )}{3} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=\frac {1}{9} \\ \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 {1}{9} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\cos \left (3 x \right ) c_{1} +c_{2} \sin \left (3 x \right )+\frac {1}{9} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +\frac {1}{9} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 \sin \left (3 x \right ) c_{1} +3 c_{2} \cos \left (3 x \right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=3 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {1}{9}, c_{2} =0\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {1}{9}-\frac {\cos \left (3 x \right )}{9} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {1}{9}-\frac {\cos \left (3 x \right )}{9} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 4.921 (sec). Leaf size: 12
\[
y \left (x \right ) = -\frac {\cos \left (3 x \right )}{9}+\frac {1}{9}
\]
✓ Solution by Mathematica
Time used: 0.021 (sec). Leaf size: 17
\[
y(x)\to \frac {2}{9} \sin ^2\left (\frac {3 x}{2}\right )
\]
14.3.2 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(x),x$2)+9*y(x)=1,y(0) = 0, D(y)(0) = 0],y(x), singsol=all)
DSolve[{y''[x]+9*y[x]==1,{y[0]==0,y'[0]==0}},y[x],x,IncludeSingularSolutions -> True]