Internal problem ID [5206]
Internal file name [OUTPUT/4699_Sunday_June_05_2022_03_03_36_PM_97048772/index.tex]
Book: Schaums Outline Differential Equations, 4th edition. Bronson and Costa. McGraw Hill
2014
Section: Chapter 24. Solutions of linear DE by Laplace transforms. Supplementary Problems.
page 248
Problem number: Problem 24.29.
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]]
\[ \boxed {y^{\prime \prime }+2 y^{\prime }-3 y=\sin \left (2 x \right )} \] 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(x)y^{\prime } + q(x) y &= F \end {align*}
Where here \begin {align*} p(x) &=2\\ q(x) &=-3\\ F &=\sin \left (2 x \right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+2 y^{\prime }-3 y = \sin \left (2 x \right ) \end {align*}
The domain of \(p(x)=2\) 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 )+2 s Y \left (s \right )-2 y \left (0\right )-3 Y \left (s \right ) = \frac {2}{s^{2}+4}\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 )+2 s Y \left (s \right )-3 Y \left (s \right ) = \frac {2}{s^{2}+4} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {2}{\left (s^{2}+4\right ) \left (s^{2}+2 s -3\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {-\frac {2}{65}+\frac {7 i}{130}}{s -2 i}+\frac {-\frac {2}{65}-\frac {7 i}{130}}{s +2 i}-\frac {1}{26 \left (s +3\right )}+\frac {1}{10 s -10} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {-\frac {2}{65}+\frac {7 i}{130}}{s -2 i}\right ) &= \left (-\frac {2}{65}+\frac {7 i}{130}\right ) {\mathrm e}^{2 i x}\\ \mathcal {L}^{-1}\left (\frac {-\frac {2}{65}-\frac {7 i}{130}}{s +2 i}\right ) &= \left (-\frac {2}{65}-\frac {7 i}{130}\right ) {\mathrm e}^{-2 i x}\\ \mathcal {L}^{-1}\left (-\frac {1}{26 \left (s +3\right )}\right ) &= -\frac {{\mathrm e}^{-3 x}}{26}\\ \mathcal {L}^{-1}\left (\frac {1}{10 s -10}\right ) &= \frac {{\mathrm e}^{x}}{10} \end {align*}
Adding the above results and simplifying gives \[ y=-\frac {4 \cos \left (2 x \right )}{65}-\frac {7 \sin \left (2 x \right )}{65}+\frac {{\mathrm e}^{x}}{10}-\frac {{\mathrm e}^{-3 x}}{26} \] Simplifying the solution gives \[
y = -\frac {4 \left (\left (\cos \left (2 x \right )+\frac {7 \sin \left (2 x \right )}{4}\right ) {\mathrm e}^{3 x}-\frac {13 \,{\mathrm e}^{4 x}}{8}+\frac {5}{8}\right ) {\mathrm e}^{-3 x}}{65}
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= -\frac {4 \left (\left (\cos \left (2 x \right )+\frac {7 \sin \left (2 x \right )}{4}\right ) {\mathrm e}^{3 x}-\frac {13 \,{\mathrm e}^{4 x}}{8}+\frac {5}{8}\right ) {\mathrm e}^{-3 x}}{65} \\
\end{align*} Verification of solutions
\[
y = -\frac {4 \left (\left (\cos \left (2 x \right )+\frac {7 \sin \left (2 x \right )}{4}\right ) {\mathrm e}^{3 x}-\frac {13 \,{\mathrm e}^{4 x}}{8}+\frac {5}{8}\right ) {\mathrm e}^{-3 x}}{65}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+2 y^{\prime }-3 y=\sin \left (2 x \right ), 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}+2 r -3=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +3\right ) \left (r -1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3, 1\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )={\mathrm e}^{-3 x} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )={\mathrm e}^{x} \\ \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=c_{1} {\mathrm e}^{-3 x}+c_{2} {\mathrm e}^{x}+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 )=\sin \left (2 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} {\mathrm e}^{-3 x} & {\mathrm e}^{x} \\ -3 \,{\mathrm e}^{-3 x} & {\mathrm e}^{x} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=4 \,{\mathrm e}^{-2 x} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & y_{p}\left (x \right )=\frac {\left ({\mathrm e}^{4 x} \left (\int {\mathrm e}^{-x} \sin \left (2 x \right )d x \right )-\left (\int \sin \left (2 x \right ) {\mathrm e}^{3 x}d x \right )\right ) {\mathrm e}^{-3 x}}{4} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=-\frac {4 \cos \left (2 x \right )}{65}-\frac {7 \sin \left (2 x \right )}{65} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-3 x}+c_{2} {\mathrm e}^{x}-\frac {4 \cos \left (2 x \right )}{65}-\frac {7 \sin \left (2 x \right )}{65} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-3 x}+c_{2} {\mathrm e}^{x}-\frac {4 \cos \left (2 x \right )}{65}-\frac {7 \sin \left (2 x \right )}{65} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} -\frac {4}{65} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 c_{1} {\mathrm e}^{-3 x}+c_{2} {\mathrm e}^{x}+\frac {8 \sin \left (2 x \right )}{65}-\frac {14 \cos \left (2 x \right )}{65} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-3 c_{1} +c_{2} -\frac {14}{65} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {1}{26}, c_{2} =\frac {1}{10}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {4 \left (\left (\cos \left (2 x \right )+\frac {7 \sin \left (2 x \right )}{4}\right ) {\mathrm e}^{3 x}-\frac {13 \,{\mathrm e}^{4 x}}{8}+\frac {5}{8}\right ) {\mathrm e}^{-3 x}}{65} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {4 \left (\left (\cos \left (2 x \right )+\frac {7 \sin \left (2 x \right )}{4}\right ) {\mathrm e}^{3 x}-\frac {13 \,{\mathrm e}^{4 x}}{8}+\frac {5}{8}\right ) {\mathrm e}^{-3 x}}{65} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.454 (sec). Leaf size: 27
\[
y \left (x \right ) = -\frac {4 \,{\mathrm e}^{-3 x} \left (\left (\cos \left (2 x \right )+\frac {7 \sin \left (2 x \right )}{4}\right ) {\mathrm e}^{3 x}-\frac {13 \,{\mathrm e}^{4 x}}{8}+\frac {5}{8}\right )}{65}
\]
✓ Solution by Mathematica
Time used: 0.109 (sec). Leaf size: 36
\[
y(x)\to \frac {1}{130} \left (-13 e^{-x}+5 e^{3 x}-14 \sin (2 x)+8 \cos (2 x)\right )
\]
5.7.2 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)+2*diff(y(x),x)-3*y(x)=sin(2*x),y(0) = 0, D(y)(0) = 0],y(x), singsol=all)
DSolve[{y''[x]-2*y'[x]-3*y[x]==Sin[2*x],{y[0]==0,y'[0]==0}},y[x],x,IncludeSingularSolutions -> True]