25.3 problem 3(c)

Internal problem ID [6507]
Internal file name [OUTPUT/5755_Sunday_June_05_2022_03_52_55_PM_56265727/index.tex]

Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 7. Laplace Transforms. Section 7.5 Problesm for review and discovery. Section A, Drill exercises. Page 309
Problem number: 3(c).
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, _linear, _nonhomogeneous]]

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

25.3.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(t)y^{\prime } + q(t) y &= F \end {align*}

Where here \begin {align*} p(t) &=2\\ q(t) &=-1\\ F &=t \,{\mathrm e}^{-t} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+2 y^{\prime }-y = t \,{\mathrm e}^{-t} \end {align*}

The domain of \(p(t)=2\) 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 ) = \frac {1}{\left (s +1\right )^{2}}\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 ) = \frac {1}{\left (s +1\right )^{2}} \end {align*}

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

Applying partial fractions decomposition results in \[ Y(s)= -\frac {1}{2 \left (s +1\right )^{2}}+\frac {3 \sqrt {2}}{8 \left (s -\sqrt {2}+1\right )}-\frac {3 \sqrt {2}}{8 \left (s +1+\sqrt {2}\right )} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {1}{2 \left (s +1\right )^{2}}\right ) &= -\frac {t \,{\mathrm e}^{-t}}{2}\\ \mathcal {L}^{-1}\left (\frac {3 \sqrt {2}}{8 \left (s -\sqrt {2}+1\right )}\right ) &= \frac {3 \sqrt {2}\, {\mathrm e}^{\sqrt {2}\, t -t}}{8}\\ \mathcal {L}^{-1}\left (-\frac {3 \sqrt {2}}{8 \left (s +1+\sqrt {2}\right )}\right ) &= -\frac {3 \sqrt {2}\, {\mathrm e}^{\frac {\left (-1-\sqrt {2}\right ) \left (-\frac {t \sqrt {2}\, \left (1+\sqrt {2}\right )}{-1-\sqrt {2}}-\frac {t \left (1+\sqrt {2}\right )}{-1-\sqrt {2}}\right )}{1+\sqrt {2}}}}{8} \end {align*}

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

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \frac {{\mathrm e}^{-t} \left (3 \sqrt {2}\, \sinh \left (\sqrt {2}\, t \right )-2 t \right )}{4} \] Verified OK.

25.3.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+2 y^{\prime }-y=t \,{\mathrm e}^{-t}, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \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 {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-2\right )\pm \left (\sqrt {8}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1-\sqrt {2}, \sqrt {2}-1\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{\left (-1-\sqrt {2}\right ) t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{\left (\sqrt {2}-1\right ) t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right )+y_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{\left (-1-\sqrt {2}\right ) t}+c_{2} {\mathrm e}^{\left (\sqrt {2}-1\right ) t}+y_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (t \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 (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (t \right )=-y_{1}\left (t \right ) \left (\int \frac {y_{2}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right )+y_{2}\left (t \right ) \left (\int \frac {y_{1}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right ), f \left (t \right )=t \,{\mathrm e}^{-t}\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\left [\begin {array}{cc} {\mathrm e}^{\left (-1-\sqrt {2}\right ) t} & {\mathrm e}^{\left (\sqrt {2}-1\right ) t} \\ \left (-1-\sqrt {2}\right ) {\mathrm e}^{\left (-1-\sqrt {2}\right ) t} & \left (\sqrt {2}-1\right ) {\mathrm e}^{\left (\sqrt {2}-1\right ) t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=2 \sqrt {2}\, {\mathrm e}^{-2 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {\sqrt {2}\, \left ({\mathrm e}^{-t \left (1+\sqrt {2}\right )} \left (\int t \,{\mathrm e}^{\sqrt {2}\, t}d t \right )-{\mathrm e}^{\left (\sqrt {2}-1\right ) t} \left (\int t \,{\mathrm e}^{-\sqrt {2}\, t}d t \right )\right )}{4} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {t \,{\mathrm e}^{-t}}{2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{\left (-1-\sqrt {2}\right ) t}+c_{2} {\mathrm e}^{\left (\sqrt {2}-1\right ) t}-\frac {t \,{\mathrm e}^{-t}}{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{\left (-1-\sqrt {2}\right ) t}+c_{2} {\mathrm e}^{\left (\sqrt {2}-1\right ) t}-\frac {t {\mathrm e}^{-t}}{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{1} \left (-1-\sqrt {2}\right ) {\mathrm e}^{\left (-1-\sqrt {2}\right ) t}+c_{2} \left (\sqrt {2}-1\right ) {\mathrm e}^{\left (\sqrt {2}-1\right ) t}-\frac {{\mathrm e}^{-t}}{2}+\frac {t \,{\mathrm e}^{-t}}{2} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=c_{1} \left (-1-\sqrt {2}\right )+c_{2} \left (\sqrt {2}-1\right )-\frac {1}{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {3 \sqrt {2}}{8}, c_{2} =\frac {3 \sqrt {2}}{8}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {3 \sqrt {2}\, {\mathrm e}^{-t \left (1+\sqrt {2}\right )}}{8}+\frac {3 \sqrt {2}\, {\mathrm e}^{\left (\sqrt {2}-1\right ) t}}{8}-\frac {t \,{\mathrm e}^{-t}}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {3 \sqrt {2}\, {\mathrm e}^{-t \left (1+\sqrt {2}\right )}}{8}+\frac {3 \sqrt {2}\, {\mathrm e}^{\left (\sqrt {2}-1\right ) t}}{8}-\frac {t \,{\mathrm e}^{-t}}{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`
 

✓ Solution by Maple

Time used: 1.969 (sec). Leaf size: 25

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

\[ y \left (t \right ) = \frac {{\mathrm e}^{-t} \left (3 \sqrt {2}\, \sinh \left (\sqrt {2}\, t \right )-2 t \right )}{4} \]

✓ Solution by Mathematica

Time used: 0.029 (sec). Leaf size: 51

DSolve[{y''[t]+2*y'[t]-y[t]==t*Exp[-t],{y[0]==0,y'[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{8} e^{-t} \left (-4 t-3 \sqrt {2} e^{-\sqrt {2} t}+3 \sqrt {2} e^{\sqrt {2} t}\right ) \]