problem 12

Internal problem ID [12803]
Internal ﬁle name [OUTPUT/11456_Saturday_November_04_2023_08_47_24_AM_94980606/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: 12.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeﬀ"

\[ \boxed {y^{\prime \prime }-y^{\prime }-2 y=x^{2}} \] With initial conditions \begin {align*} \left [y \left (0\right ) = {\frac {11}{4}}, y^{\prime }\left (0\right ) = {\frac {1}{2}}\right ] \end {align*}

14.6.1 Existence and uniqueness analysis

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*}

Hence the ode is \begin {align*} y^{\prime \prime }-y^{\prime }-2 y = x^{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 )-s Y \left (s \right )+y \left (0\right )-2 Y \left (s \right ) = \frac {2}{s^{3}}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&={\frac {11}{4}}\\ y'(0) &={\frac {1}{2}} \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+\frac {9}{4}-\frac {11 s}{4}-s Y \left (s \right )-2 Y \left (s \right ) = \frac {2}{s^{3}} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {11 s^{4}-9 s^{3}+8}{4 s^{3} \left (s^{2}-s -2\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {7}{3 \left (s +1\right )}+\frac {7}{6 \left (s -2\right )}-\frac {3}{4 s}-\frac {1}{s^{3}}+\frac {1}{2 s^{2}} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {7}{3 \left (s +1\right )}\right ) &= \frac {7 \,{\mathrm e}^{-x}}{3}\\ \mathcal {L}^{-1}\left (\frac {7}{6 \left (s -2\right )}\right ) &= \frac {7 \,{\mathrm e}^{2 x}}{6}\\ \mathcal {L}^{-1}\left (-\frac {3}{4 s}\right ) &= -{\frac {3}{4}}\\ \mathcal {L}^{-1}\left (-\frac {1}{s^{3}}\right ) &= -\frac {x^{2}}{2}\\ \mathcal {L}^{-1}\left (\frac {1}{2 s^{2}}\right ) &= \frac {x}{2} \end {align*}

Adding the above results and simplifying gives \[ y=-\frac {3}{4}-\frac {x^{2}}{2}+\frac {7 \,{\mathrm e}^{-x}}{3}+\frac {7 \,{\mathrm e}^{2 x}}{6}+\frac {x}{2} \] Simplifying the solution gives \[ y = -\frac {3}{4}-\frac {x^{2}}{2}+\frac {7 \,{\mathrm e}^{-x}}{3}+\frac {7 \,{\mathrm e}^{2 x}}{6}+\frac {x}{2} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {3}{4}-\frac {x^{2}}{2}+\frac {7 \,{\mathrm e}^{-x}}{3}+\frac {7 \,{\mathrm e}^{2 x}}{6}+\frac {x}{2} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -\frac {3}{4}-\frac {x^{2}}{2}+\frac {7 \,{\mathrm e}^{-x}}{3}+\frac {7 \,{\mathrm e}^{2 x}}{6}+\frac {x}{2} \] Veriﬁed OK.

14.6.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-y^{\prime }-2 y=x^{2}, y \left (0\right )=\frac {11}{4}, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=\frac {1}{2}\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}-r -2=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +1\right ) \left (r -2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1, 2\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )={\mathrm e}^{-x} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )={\mathrm e}^{2 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={\mathrm e}^{-x} c_{1} +c_{2} {\mathrm e}^{2 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 )=x^{2}\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}^{-x} & {\mathrm e}^{2 x} \\ -{\mathrm e}^{-x} & 2 \,{\mathrm e}^{2 x} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=3 \,{\mathrm e}^{x} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & y_{p}\left (x \right )=-\frac {{\mathrm e}^{-x} \left (\int {\mathrm e}^{x} x^{2}d x \right )}{3}+\frac {{\mathrm e}^{2 x} \left (\int {\mathrm e}^{-2 x} x^{2}d x \right )}{3} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=-\frac {1}{2} x^{2}+\frac {1}{2} x -\frac {3}{4} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-x} c_{1} +c_{2} {\mathrm e}^{2 x}-\frac {x^{2}}{2}+\frac {x}{2}-\frac {3}{4} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y={\mathrm e}^{-x} c_{1} +c_{2} {\mathrm e}^{2 x}-\frac {x^{2}}{2}+\frac {x}{2}-\frac {3}{4} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=\frac {11}{4} \\ {} & {} & \frac {11}{4}=c_{1} +c_{2} -\frac {3}{4} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-{\mathrm e}^{-x} c_{1} +2 c_{2} {\mathrm e}^{2 x}-x +\frac {1}{2} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=\frac {1}{2} \\ {} & {} & \frac {1}{2}=-c_{1} +2 c_{2} +\frac {1}{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {7}{3}, c_{2} =\frac {7}{6}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {3}{4}-\frac {x^{2}}{2}+\frac {7 \,{\mathrm e}^{-x}}{3}+\frac {7 \,{\mathrm e}^{2 x}}{6}+\frac {x}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {3}{4}-\frac {x^{2}}{2}+\frac {7 \,{\mathrm e}^{-x}}{3}+\frac {7 \,{\mathrm e}^{2 x}}{6}+\frac {x}{2} \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`

\[ y \left (x \right ) = \frac {7 \,{\mathrm e}^{-x}}{3}+\frac {x}{2}-\frac {x^{2}}{2}+\frac {7 \,{\mathrm e}^{2 x}}{6}-\frac {3}{4} \]

14.6 problem 12

14.6.1 Existence and uniqueness analysis

14.6.2 Maple step by step solution