Internal problem ID [12785]
Internal file name [OUTPUT/11438_Saturday_November_04_2023_08_47_19_AM_72649872/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: 2.
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]]
\[ \boxed {y^{\prime \prime }-2 y^{\prime }+5 y=0} \] 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 )-2 s Y \left (s \right )+2 y \left (0\right )+5 Y \left (s \right ) = 0\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} -2 s Y \left (s \right )+2 c_{1} +5 Y \left (s \right ) = 0 \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s c_{1} -2 c_{1} +c_{2}}{s^{2}-2 s +5} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {\left (1+2 i\right ) \left (\frac {c_{1}}{8}-\frac {c_{2}}{8}\right )+\frac {3 c_{1}}{8}+\frac {c_{2}}{8}}{s -1-2 i}+\frac {\left (1-2 i\right ) \left (\frac {c_{1}}{8}-\frac {c_{2}}{8}\right )+\frac {3 c_{1}}{8}+\frac {c_{2}}{8}}{s -1+2 i} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {\left (1+2 i\right ) \left (\frac {c_{1}}{8}-\frac {c_{2}}{8}\right )+\frac {3 c_{1}}{8}+\frac {c_{2}}{8}}{s -1-2 i}\right ) &= \frac {{\mathrm e}^{\left (1+2 i\right ) x} \left (-i c_{2} +\left (2+i\right ) c_{1} \right )}{4}\\ \mathcal {L}^{-1}\left (\frac {\left (1-2 i\right ) \left (\frac {c_{1}}{8}-\frac {c_{2}}{8}\right )+\frac {3 c_{1}}{8}+\frac {c_{2}}{8}}{s -1+2 i}\right ) &= \frac {{\mathrm e}^{\left (1-2 i\right ) x} \left (i c_{2} +\left (2-i\right ) c_{1} \right )}{4} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {{\mathrm e}^{x} \left (2 c_{1} \cos \left (2 x \right )+\sin \left (2 x \right ) \left (-c_{1} +c_{2} \right )\right )}{2} \] Simplifying the solution gives \[ y = \frac {\left (-c_{1} +c_{2} \right ) {\mathrm e}^{x} \sin \left (2 x \right )}{2}+c_{1} \cos \left (2 x \right ) {\mathrm e}^{x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-c_{1} +c_{2} \right ) {\mathrm e}^{x} \sin \left (2 x \right )}{2}+c_{1} \cos \left (2 x \right ) {\mathrm e}^{x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-c_{1} +c_{2} \right ) {\mathrm e}^{x} \sin \left (2 x \right )}{2}+c_{1} \cos \left (2 x \right ) {\mathrm e}^{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-2 y^{\prime }+5 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-2 r +5=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {2\pm \left (\sqrt {-16}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (1-2 \,\mathrm {I}, 1+2 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )={\mathrm e}^{x} \cos \left (2 x \right ) \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )={\mathrm e}^{x} \sin \left (2 x \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (2 x \right ) {\mathrm e}^{x}+c_{2} {\mathrm e}^{x} \sin \left (2 x \right ) \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 5.516 (sec). Leaf size: 29
dsolve(diff(y(x),x$2)-2*diff(y(x),x)+5*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{x} \left (2 y \left (0\right ) \cos \left (2 x \right )+\sin \left (2 x \right ) \left (D\left (y \right )\left (0\right )-y \left (0\right )\right )\right )}{2} \]
✓ Solution by Mathematica
Time used: 0.024 (sec). Leaf size: 24
DSolve[y''[x]-2*y'[x]+5*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^x (c_2 \cos (2 x)+c_1 \sin (2 x)) \]