5.6 problem Problem 24.28

Internal problem ID [5205]
Internal file name [OUTPUT/4698_Sunday_June_05_2022_03_03_35_PM_85166985/index.tex]

Book: Schaums Outline Differential Equations, 4th edition. Bronson and Costa. McGraw Hill 2014
Section: Chapter 24. Solutions of linear DE by Laplace transforms. Supplementary Problems. page 248
Problem number: Problem 24.28.
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]]

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

5.6.1 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*}

Where here \begin {align*} p(x) &=0\\ q(x) &=-1\\ F &={\mathrm e}^{x} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-y = {\mathrm e}^{x} \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 )-Y \left (s \right ) = \frac {1}{s -1}\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 )-s -Y \left (s \right ) = \frac {1}{s -1} \end {align*}

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

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

Adding the above results and simplifying gives \[ y=\frac {3 \,{\mathrm e}^{-x}}{4}+\frac {{\mathrm e}^{x} \left (2 x +1\right )}{4} \] Simplifying the solution gives \[ y = \frac {3 \,{\mathrm e}^{-x}}{4}+\frac {{\mathrm e}^{x} \left (2 x +1\right )}{4} \]

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \frac {3 \,{\mathrm e}^{-x}}{4}+\frac {{\mathrm e}^{x} \left (2 x +1\right )}{4} \] Verified OK.

5.6.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-y={\mathrm e}^{x}, 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}-1=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -1\right ) \left (r +1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1, 1\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}^{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} {\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 )={\mathrm e}^{x}\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}^{x} \\ -{\mathrm e}^{-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 )=2 \\ {} & \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}^{2 x}d x \right )}{2}+\frac {{\mathrm e}^{x} \left (\int 1d x \right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=\frac {\left (2 x -1\right ) {\mathrm e}^{x}}{4} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+\frac {\left (2 x -1\right ) {\mathrm e}^{x}}{4} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+\frac {\left (2 x -1\right ) {\mathrm e}^{x}}{4} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} +c_{2} -\frac {1}{4} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+\frac {{\mathrm e}^{x}}{2}+\frac {\left (2 x -1\right ) {\mathrm e}^{x}}{4} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-c_{1} +c_{2} +\frac {1}{4} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {3}{4}, c_{2} =\frac {1}{2}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {3 \,{\mathrm e}^{-x}}{4}+\frac {{\mathrm e}^{x} \left (2 x +1\right )}{4} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {3 \,{\mathrm e}^{-x}}{4}+\frac {{\mathrm e}^{x} \left (2 x +1\right )}{4} \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: 0.391 (sec). Leaf size: 20

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

\[ y \left (x \right ) = \frac {3 \,{\mathrm e}^{-x}}{4}+\frac {{\mathrm e}^{x} \left (2 x +1\right )}{4} \]

✓ Solution by Mathematica

Time used: 0.017 (sec). Leaf size: 27

DSolve[{y''[x]-y[x]==Exp[x],{y[0]==1,y'[0]==0}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{4} e^{-x} \left (e^{2 x} (2 x+1)+3\right ) \]