problem 13

Internal problem ID [12796]
Internal ﬁle name [OUTPUT/11449_Saturday_November_04_2023_08_47_22_AM_10331552/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: 13.
ODE order: 2.
ODE degree: 1.

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

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

13.13.1 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 }-2 y^{\prime }+2 y = -x^{2}+1 \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 )+2 Y \left (s \right ) = -\frac {2}{s^{3}}+\frac {1}{s}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=1\\ 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 -2 s Y \left (s \right )+2 Y \left (s \right ) = -\frac {2}{s^{3}}+\frac {1}{s} \end {align*}

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

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

Adding the above results and simplifying gives \[ y={\mathrm e}^{x} \cos \left (x \right )-\frac {x^{2}}{2}-x \] Simplifying the solution gives \[ y = {\mathrm e}^{x} \cos \left (x \right )-\frac {x^{2}}{2}-x \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{x} \cos \left (x \right )-\frac {x^{2}}{2}-x \\ \end{align*}

Veriﬁcation of solutions

\[ y = {\mathrm e}^{x} \cos \left (x \right )-\frac {x^{2}}{2}-x \] Veriﬁed OK.

13.13.2 Maple step by step solution

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

\[ y \left (x \right ) = -x -\frac {x^{2}}{2}+\cos \left (x \right ) {\mathrm e}^{x} \]

13.13 problem 13

13.13.1 Existence and uniqueness analysis

13.13.2 Maple step by step solution