Internal problem ID [6512]
Internal file name [OUTPUT/5760_Sunday_June_05_2022_03_53_04_PM_51888636/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: 4(d).
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]]
\[ \boxed {y^{\prime \prime }-7 y^{\prime }+12 y=t \,{\mathrm e}^{2 t}} \] Since no initial conditions are explicitly given, then let \begin {align*} y \left (0\right )&= c_{1} \\ y'(0) &= c_{2} \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 )-7 s Y \left (s \right )+7 y \left (0\right )+12 Y \left (s \right ) = \frac {1}{\left (s -2\right )^{2}}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=c_{1}\\ y'(0) &=c_{2} \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-c_{2} -s c_{1} -7 s Y \left (s \right )+7 c_{1} +12 Y \left (s \right ) = \frac {1}{\left (s -2\right )^{2}} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {c_{1} s^{3}-11 c_{1} s^{2}+c_{2} s^{2}+32 s c_{1} -4 c_{2} s -28 c_{1} +4 c_{2} +1}{\left (s -2\right )^{2} \left (s^{2}-7 s +12\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {-3 c_{1} +c_{2} +\frac {1}{4}}{s -4}+\frac {4 c_{1} -c_{2} -1}{s -3}+\frac {1}{2 \left (s -2\right )^{2}}+\frac {3}{4 \left (s -2\right )} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {-3 c_{1} +c_{2} +\frac {1}{4}}{s -4}\right ) &= \frac {{\mathrm e}^{4 t} \left (-12 c_{1} +4 c_{2} +1\right )}{4}\\ \mathcal {L}^{-1}\left (\frac {4 c_{1} -c_{2} -1}{s -3}\right ) &= \left (4 c_{1} -c_{2} -1\right ) {\mathrm e}^{3 t}\\ \mathcal {L}^{-1}\left (\frac {1}{2 \left (s -2\right )^{2}}\right ) &= \frac {t \,{\mathrm e}^{2 t}}{2}\\ \mathcal {L}^{-1}\left (\frac {3}{4 \left (s -2\right )}\right ) &= \frac {3 \,{\mathrm e}^{2 t}}{4} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {{\mathrm e}^{2 t} \left (2 t +3\right )}{4}+\left (4 c_{1} -c_{2} -1\right ) {\mathrm e}^{3 t}+\frac {{\mathrm e}^{4 t} \left (-12 c_{1} +4 c_{2} +1\right )}{4} \] Simplifying the solution gives \[ y = \frac {{\mathrm e}^{2 t} \left (2 t +3\right )}{4}+\left (4 c_{1} -c_{2} -1\right ) {\mathrm e}^{3 t}+\frac {{\mathrm e}^{4 t} \left (-12 c_{1} +4 c_{2} +1\right )}{4} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{2 t} \left (2 t +3\right )}{4}+\left (4 c_{1} -c_{2} -1\right ) {\mathrm e}^{3 t}+\frac {{\mathrm e}^{4 t} \left (-12 c_{1} +4 c_{2} +1\right )}{4} \\ \end{align*}
Verification of solutions
\[ y = \frac {{\mathrm e}^{2 t} \left (2 t +3\right )}{4}+\left (4 c_{1} -c_{2} -1\right ) {\mathrm e}^{3 t}+\frac {{\mathrm e}^{4 t} \left (-12 c_{1} +4 c_{2} +1\right )}{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-7 y^{\prime }+12 y=t \,{\mathrm e}^{2 t} \\ \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}-7 r +12=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -3\right ) \left (r -4\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (3, 4\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{3 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{4 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={\mathrm e}^{3 t} c_{1} +c_{2} {\mathrm e}^{4 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}^{2 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}^{3 t} & {\mathrm e}^{4 t} \\ 3 \,{\mathrm e}^{3 t} & 4 \,{\mathrm e}^{4 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )={\mathrm e}^{7 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-{\mathrm e}^{3 t} \left (\int t \,{\mathrm e}^{-t}d t \right )+{\mathrm e}^{4 t} \left (\int t \,{\mathrm e}^{-2 t}d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {{\mathrm e}^{2 t} \left (2 t +3\right )}{4} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{3 t} c_{1} +c_{2} {\mathrm e}^{4 t}+\frac {{\mathrm e}^{2 t} \left (2 t +3\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: 1.828 (sec). Leaf size: 47
dsolve(diff(y(t),t$2)-7*diff(y(t),t)+12*y(t)=t*exp(2*t),y(t), singsol=all)
\[ y \left (t \right ) = \frac {\left (2 t +3\right ) {\mathrm e}^{2 t}}{4}+\left (4 y \left (0\right )-D\left (y \right )\left (0\right )-1\right ) {\mathrm e}^{3 t}+\frac {{\mathrm e}^{4 t} \left (-12 y \left (0\right )+4 D\left (y \right )\left (0\right )+1\right )}{4} \]
✓ Solution by Mathematica
Time used: 0.019 (sec). Leaf size: 35
DSolve[y''[t]-7*y'[t]+12*y[t]==t*Exp[2*t],y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {1}{4} e^{2 t} \left (2 t+4 c_1 e^t+4 c_2 e^{2 t}+3\right ) \]