13.10 problem 10

Internal problem ID [12793]
Internal file name [OUTPUT/11446_Saturday_November_04_2023_08_47_21_AM_34637621/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: 10.
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, _with_linear_symmetries]]

\[ \boxed {y^{\prime \prime }-9 y=x +2} \] With initial conditions \begin {align*} [y \left (0\right ) = -1, y^{\prime }\left (0\right ) = 1] \end {align*}

13.10.1 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*}

Where here \begin {align*} p(x) &=0\\ q(x) &=-9\\ F &=x +2 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-9 y = x +2 \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 {2 s +1}{s^{2}}\tag {1} \end {align*}

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

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

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {s^{3}-s^{2}-2 s -1}{s^{2} \left (s^{2}-9\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= -\frac {11}{54 \left (s -3\right )}-\frac {2}{9 s}-\frac {31}{54 \left (s +3\right )}-\frac {1}{9 s^{2}} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {11}{54 \left (s -3\right )}\right ) &= -\frac {11 \,{\mathrm e}^{3 x}}{54}\\ \mathcal {L}^{-1}\left (-\frac {2}{9 s}\right ) &= -{\frac {2}{9}}\\ \mathcal {L}^{-1}\left (-\frac {31}{54 \left (s +3\right )}\right ) &= -\frac {31 \,{\mathrm e}^{-3 x}}{54}\\ \mathcal {L}^{-1}\left (-\frac {1}{9 s^{2}}\right ) &= -\frac {x}{9} \end {align*}

Adding the above results and simplifying gives \[ y=-\frac {2}{9}-\frac {x}{9}-\frac {7 \cosh \left (3 x \right )}{9}+\frac {10 \sinh \left (3 x \right )}{27} \] Simplifying the solution gives \[ y = -\frac {2}{9}-\frac {x}{9}-\frac {7 \cosh \left (3 x \right )}{9}+\frac {10 \sinh \left (3 x \right )}{27} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {2}{9}-\frac {x}{9}-\frac {7 \cosh \left (3 x \right )}{9}+\frac {10 \sinh \left (3 x \right )}{27} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -\frac {2}{9}-\frac {x}{9}-\frac {7 \cosh \left (3 x \right )}{9}+\frac {10 \sinh \left (3 x \right )}{27} \] Verified OK.

13.10.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-9 y=x +2, y \left (0\right )=-1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1\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 {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -3\right ) \left (r +3\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3, 3\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}^{3 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}^{3 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}^{-3 x} & {\mathrm e}^{3 x} \\ -3 \,{\mathrm e}^{-3 x} & 3 \,{\mathrm e}^{3 x} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=6 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & y_{p}\left (x \right )=-\frac {{\mathrm e}^{-3 x} \left (\int {\mathrm e}^{3 x} \left (x +2\right )d x \right )}{6}+\frac {{\mathrm e}^{3 x} \left (\int {\mathrm e}^{-3 x} \left (x +2\right )d x \right )}{6} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=-\frac {2}{9}-\frac {x}{9} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-3 x}+c_{2} {\mathrm e}^{3 x}-\frac {2}{9}-\frac {x}{9} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-3 x}+c_{2} {\mathrm e}^{3 x}-\frac {2}{9}-\frac {x}{9} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-1 \\ {} & {} & -1=c_{1} +c_{2} -\frac {2}{9} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 c_{1} {\mathrm e}^{-3 x}+3 c_{2} {\mathrm e}^{3 x}-\frac {1}{9} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-3 c_{1} +3 c_{2} -\frac {1}{9} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {31}{54}, c_{2} =-\frac {11}{54}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {31 \,{\mathrm e}^{-3 x}}{54}-\frac {11 \,{\mathrm e}^{3 x}}{54}-\frac {2}{9}-\frac {x}{9} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {31 \,{\mathrm e}^{-3 x}}{54}-\frac {11 \,{\mathrm e}^{3 x}}{54}-\frac {2}{9}-\frac {x}{9} \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: 5.391 (sec). Leaf size: 21

dsolve([diff(y(x),x$2)-9*y(x)=x+2,y(0) = -1, D(y)(0) = 1],y(x), singsol=all)
 

\[ y \left (x \right ) = -\frac {x}{9}-\frac {7 \cosh \left (3 x \right )}{9}+\frac {10 \sinh \left (3 x \right )}{27}-\frac {2}{9} \]

✓ Solution by Mathematica

Time used: 0.026 (sec). Leaf size: 33

DSolve[{y''[x]-9*y[x]==x+2,{y[0]==-1,y'[0]==1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{54} e^{-3 x} \left (-6 e^{3 x} (x+2)-11 e^{6 x}-31\right ) \]