Internal problem ID [12319]
Internal file name [OUTPUT/10972_Monday_October_02_2023_02_47_39_AM_45017633/index.tex]
Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A.
Dobrushkin. CRC Press 2015
Section: Chapter 5.6 Laplace transform. Nonhomogeneous equations. Problems page
368
Problem number: Problem 2(k)[l].
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]]
\[ \boxed {2 y^{\prime \prime }+y^{\prime }-y=4 \sin \left (t \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = -4] \end {align*}
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) &={\frac {1}{2}}\\ q(t) &=-{\frac {1}{2}}\\ F &=2 \sin \left (t \right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+\frac {y^{\prime }}{2}-\frac {y}{2} = 2 \sin \left (t \right ) \end {align*}
The domain of \(p(t)={\frac {1}{2}}\) 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*} 2 s^{2} Y \left (s \right )-2 y^{\prime }\left (0\right )-2 s y \left (0\right )+s Y \left (s \right )-y \left (0\right )-Y \left (s \right ) = \frac {4}{s^{2}+1}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y'(0) &=-4 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} 2 s^{2} Y \left (s \right )+8+s Y \left (s \right )-Y \left (s \right ) = \frac {4}{s^{2}+1} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {4 \left (2 s^{2}+1\right )}{\left (s^{2}+1\right ) \left (2 s^{2}+s -1\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {-\frac {1}{5}+\frac {3 i}{5}}{s -i}+\frac {-\frac {1}{5}-\frac {3 i}{5}}{s +i}+\frac {2}{s +1}-\frac {8}{5 \left (s -\frac {1}{2}\right )} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {-\frac {1}{5}+\frac {3 i}{5}}{s -i}\right ) &= \left (-\frac {1}{5}+\frac {3 i}{5}\right ) {\mathrm e}^{i t}\\ \mathcal {L}^{-1}\left (\frac {-\frac {1}{5}-\frac {3 i}{5}}{s +i}\right ) &= \left (-\frac {1}{5}-\frac {3 i}{5}\right ) {\mathrm e}^{-i t}\\ \mathcal {L}^{-1}\left (\frac {2}{s +1}\right ) &= 2 \,{\mathrm e}^{-t}\\ \mathcal {L}^{-1}\left (-\frac {8}{5 \left (s -\frac {1}{2}\right )}\right ) &= -\frac {8 \,{\mathrm e}^{\frac {t}{2}}}{5} \end {align*}
Adding the above results and simplifying gives \[ y=2 \,{\mathrm e}^{-t}-\frac {8 \,{\mathrm e}^{\frac {t}{2}}}{5}-\frac {2 \cos \left (t \right )}{5}-\frac {6 \sin \left (t \right )}{5} \] Simplifying the solution gives \[
y = -\frac {2 \left (4 \,{\mathrm e}^{\frac {3 t}{2}}-5+\left (\cos \left (t \right )+3 \sin \left (t \right )\right ) {\mathrm e}^{t}\right ) {\mathrm e}^{-t}}{5}
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= -\frac {2 \left (4 \,{\mathrm e}^{\frac {3 t}{2}}-5+\left (\cos \left (t \right )+3 \sin \left (t \right )\right ) {\mathrm e}^{t}\right ) {\mathrm e}^{-t}}{5} \\
\end{align*} Verification of solutions
\[
y = -\frac {2 \left (4 \,{\mathrm e}^{\frac {3 t}{2}}-5+\left (\cos \left (t \right )+3 \sin \left (t \right )\right ) {\mathrm e}^{t}\right ) {\mathrm e}^{-t}}{5}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [2 y^{\prime \prime }+y^{\prime }-y=4 \sin \left (t \right ), y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-4\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {y^{\prime }}{2}+\frac {y}{2}+2 \sin \left (t \right ) \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {y^{\prime }}{2}-\frac {y}{2}=2 \sin \left (t \right ) \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+\frac {1}{2} r -\frac {1}{2}=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \frac {\left (r +1\right ) \left (2 r -1\right )}{2}=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1, \frac {1}{2}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{\frac {t}{2}} \\ \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}^{-t}+c_{2} {\mathrm e}^{\frac {t}{2}}+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 )=2 \sin \left (t \right )\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}^{-t} & {\mathrm e}^{\frac {t}{2}} \\ -{\mathrm e}^{-t} & \frac {{\mathrm e}^{\frac {t}{2}}}{2} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\frac {3 \,{\mathrm e}^{-\frac {t}{2}}}{2} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=\frac {4 \left ({\mathrm e}^{\frac {3 t}{2}} \left (\int \sin \left (t \right ) {\mathrm e}^{-\frac {t}{2}}d t \right )-\left (\int \sin \left (t \right ) {\mathrm e}^{t}d t \right )\right ) {\mathrm e}^{-t}}{3} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {2 \cos \left (t \right )}{5}-\frac {6 \sin \left (t \right )}{5} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{\frac {t}{2}}-\frac {2 \cos \left (t \right )}{5}-\frac {6 \sin \left (t \right )}{5} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{\frac {t}{2}}-\frac {2 \cos \left (t \right )}{5}-\frac {6 \sin \left (t \right )}{5} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} -\frac {2}{5} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} {\mathrm e}^{-t}+\frac {c_{2} {\mathrm e}^{\frac {t}{2}}}{2}+\frac {2 \sin \left (t \right )}{5}-\frac {6 \cos \left (t \right )}{5} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-4 \\ {} & {} & -4=-c_{1} +\frac {c_{2}}{2}-\frac {6}{5} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =2, c_{2} =-\frac {8}{5}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {2 \left (4 \,{\mathrm e}^{\frac {3 t}{2}}-5+\left (\cos \left (t \right )+3 \sin \left (t \right )\right ) {\mathrm e}^{t}\right ) {\mathrm e}^{-t}}{5} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {2 \left (4 \,{\mathrm e}^{\frac {3 t}{2}}-5+\left (\cos \left (t \right )+3 \sin \left (t \right )\right ) {\mathrm e}^{t}\right ) {\mathrm e}^{-t}}{5} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 4.516 (sec). Leaf size: 25
\[
y \left (t \right ) = -\frac {2 \,{\mathrm e}^{-t} \left (4 \,{\mathrm e}^{\frac {3 t}{2}}-5+\left (\cos \left (t \right )+3 \sin \left (t \right )\right ) {\mathrm e}^{t}\right )}{5}
\]
✓ Solution by Mathematica
Time used: 0.037 (sec). Leaf size: 34
\[
y(t)\to \frac {2}{5} \left (5 e^{-t}-4 e^{t/2}-3 \sin (t)-\cos (t)\right )
\]
4.12.2 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([2*diff(y(t),t$2)+diff(y(t),t)-y(t)=4*sin(t),y(0) = 0, D(y)(0) = -4],y(t), singsol=all)
DSolve[{2*y''[t]+y'[t]-y[t]==4*Sin[t],{y[0]==0,y'[0]==-4}},y[t],t,IncludeSingularSolutions -> True]