Internal problem ID [12816]
Internal file name [OUTPUT/11469_Saturday_November_04_2023_08_47_30_AM_42828604/index.tex]
Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 5. The Laplace Transform Method. Exercises 5.5, page 273
Problem number: 4.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeff", "linear_second_order_ode_solved_by_an_integrating_factor"
Maple gives the following as the ode type
[[_2nd_order, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime }-2 y^{\prime }+y=2 \delta \left (x -1\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \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) &=1\\ F &=2 \delta \left (x -1\right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-2 y^{\prime }+y = 2 \delta \left (x -1\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 )+Y \left (s \right ) = 2 \,{\mathrm e}^{-s}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y'(0) &=1 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-1-2 s Y \left (s \right )+Y \left (s \right ) = 2 \,{\mathrm e}^{-s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {2 \,{\mathrm e}^{-s}+1}{s^{2}-2 s +1} \end {align*}
Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {2 \,{\mathrm e}^{-s}+1}{s^{2}-2 s +1}\right )\\ &= x \,{\mathrm e}^{x}+2 \left (1-\operatorname {Heaviside}\left (1-x \right )\right ) {\mathrm e}^{x -1} \left (x -1\right ) \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} x \,{\mathrm e}^{x} & x \le 1 \\ x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x -1} \left (x -1\right ) & 1 The solution(s) found are the following \begin{align*}
\tag{1} y &= x \,{\mathrm e}^{x}+\left (\left \{\begin {array}{cc} 0 & x \le 1 \\ 2 \,{\mathrm e}^{x -1} \left (x -1\right ) & 1 Verification of solutions
\[
y = x \,{\mathrm e}^{x}+\left (\left \{\begin {array}{cc} 0 & x \le 1 \\ 2 \,{\mathrm e}^{x -1} \left (x -1\right ) & 1
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-2 y^{\prime }+y=2 \mathit {Dirac}\left (x -1\right ), y \left (0\right )=0, 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}-2 r +1=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -1\right )^{2}=0 \\ \bullet & {} & \textrm {Root of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =1 \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )={\mathrm e}^{x} \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (x \right )\hspace {3pt}\textrm {by}\hspace {3pt} x \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=x \,{\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}^{x}+c_{2} x \,{\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 )=2 \mathit {Dirac}\left (x -1\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}^{x} & x \,{\mathrm e}^{x} \\ {\mathrm e}^{x} & {\mathrm e}^{x}+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 )={\mathrm e}^{2 x} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & y_{p}\left (x \right )=2 \left (\int \mathit {Dirac}\left (x -1\right )d x \right ) {\mathrm e}^{x -1} \left (x -1\right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=2 \mathit {Heaviside}\left (x -1\right ) {\mathrm e}^{x -1} \left (x -1\right ) \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{x}+c_{2} x \,{\mathrm e}^{x}+2 \mathit {Heaviside}\left (x -1\right ) {\mathrm e}^{x -1} \left (x -1\right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{x}+c_{2} x {\mathrm e}^{x}+2 \mathit {Heaviside}\left (x -1\right ) {\mathrm e}^{x -1} \left (x -1\right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{x}+c_{2} x \,{\mathrm e}^{x}+2 \mathit {Dirac}\left (x -1\right ) {\mathrm e}^{x -1} \left (x -1\right )+2 \mathit {Heaviside}\left (x -1\right ) {\mathrm e}^{x -1} \left (x -1\right )+2 \mathit {Heaviside}\left (x -1\right ) {\mathrm e}^{x -1} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=c_{1} +c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=x \,{\mathrm e}^{x}+2 \mathit {Heaviside}\left (x -1\right ) {\mathrm e}^{x -1} \left (x -1\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=x \,{\mathrm e}^{x}+2 \mathit {Heaviside}\left (x -1\right ) {\mathrm e}^{x -1} \left (x -1\right ) \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 6.187 (sec). Leaf size: 28
\[
y \left (x \right ) = 2 \operatorname {Heaviside}\left (-1+x \right ) {\mathrm e}^{-1+x} \left (-1+x \right )+{\mathrm e}^{x} x
\]
✓ Solution by Mathematica
Time used: 0.039 (sec). Leaf size: 24
\[
y(x)\to e^{x-1} (2 (x-1) \theta (x-1)+e x)
\]
16.4.2 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)+y(x)=2*Dirac(x-1),y(0) = 0, D(y)(0) = 1],y(x), singsol=all)
DSolve[{y''[x]-2*y'[x]+y[x]==2*DiracDelta[x-1],{y[0]==0,y'[0]==1}},y[x],x,IncludeSingularSolutions -> True]