Internal problem ID [14672]
Internal file name [OUTPUT/14352_Wednesday_April_03_2024_02_17_27_PM_12648957/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.5, page 175
Problem number: 63 (b).
ODE order: 4.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_linear_constant_coefficients_ODE"
Maple gives the following as the ode type
[[_high_order, _missing_x]]
\[ \boxed {y^{\prime \prime \prime \prime }-8 y^{\prime \prime \prime }+30 y^{\prime \prime }-56 y^{\prime }+49 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 2, y^{\prime \prime }\left (0\right ) = -1, y^{\prime \prime \prime }\left (0\right ) = -1] \end {align*}
The characteristic equation is \[ \lambda ^{4}-8 \lambda ^{3}+30 \lambda ^{2}-56 \lambda +49 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= i \sqrt {3}+2\\ \lambda _2 &= -i \sqrt {3}+2\\ \lambda _3 &= i \sqrt {3}+2\\ \lambda _4 &= -i \sqrt {3}+2 \end {align*}
Therefore the homogeneous solution is \[ y_h(t)={\mathrm e}^{\left (-i \sqrt {3}+2\right ) t} c_{1} +t \,{\mathrm e}^{\left (-i \sqrt {3}+2\right ) t} c_{2} +{\mathrm e}^{\left (i \sqrt {3}+2\right ) t} c_{3} +t \,{\mathrm e}^{\left (i \sqrt {3}+2\right ) t} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{\left (-i \sqrt {3}+2\right ) t}\\ y_2 &= t \,{\mathrm e}^{\left (-i \sqrt {3}+2\right ) t}\\ y_3 &= {\mathrm e}^{\left (i \sqrt {3}+2\right ) t}\\ y_4 &= t \,{\mathrm e}^{\left (i \sqrt {3}+2\right ) t} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = {\mathrm e}^{\left (-i \sqrt {3}+2\right ) t} c_{1} +t \,{\mathrm e}^{\left (-i \sqrt {3}+2\right ) t} c_{2} +{\mathrm e}^{\left (i \sqrt {3}+2\right ) t} c_{3} +t \,{\mathrm e}^{\left (i \sqrt {3}+2\right ) t} c_{4} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 1\) and \(t = 0\) in the above gives \begin {align*} 1 = c_{1} +c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \left (-i \sqrt {3}+2\right ) {\mathrm e}^{\left (-i \sqrt {3}+2\right ) t} c_{1} +{\mathrm e}^{\left (-i \sqrt {3}+2\right ) t} c_{2} +t \left (-i \sqrt {3}+2\right ) {\mathrm e}^{\left (-i \sqrt {3}+2\right ) t} c_{2} +\left (i \sqrt {3}+2\right ) {\mathrm e}^{\left (i \sqrt {3}+2\right ) t} c_{3} +{\mathrm e}^{\left (i \sqrt {3}+2\right ) t} c_{4} +t \left (i \sqrt {3}+2\right ) {\mathrm e}^{\left (i \sqrt {3}+2\right ) t} c_{4} \end {align*}
substituting \(y^{\prime } = 2\) and \(t = 0\) in the above gives \begin {align*} 2 = i \left (-c_{1} +c_{3} \right ) \sqrt {3}+2 c_{1} +c_{2} +2 c_{3} +c_{4}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = \left (-i \sqrt {3}+2\right )^{2} {\mathrm e}^{\left (-i \sqrt {3}+2\right ) t} c_{1} +2 \left (-i \sqrt {3}+2\right ) {\mathrm e}^{\left (-i \sqrt {3}+2\right ) t} c_{2} +t \left (-i \sqrt {3}+2\right )^{2} {\mathrm e}^{\left (-i \sqrt {3}+2\right ) t} c_{2} +\left (i \sqrt {3}+2\right )^{2} {\mathrm e}^{\left (i \sqrt {3}+2\right ) t} c_{3} +2 \left (i \sqrt {3}+2\right ) {\mathrm e}^{\left (i \sqrt {3}+2\right ) t} c_{4} +t \left (i \sqrt {3}+2\right )^{2} {\mathrm e}^{\left (i \sqrt {3}+2\right ) t} c_{4} \end {align*}
substituting \(y^{\prime \prime } = -1\) and \(t = 0\) in the above gives \begin {align*} -1 = 2 i \left (-2 c_{1} -c_{2} +2 c_{3} +c_{4} \right ) \sqrt {3}+c_{1} +4 c_{2} +c_{3} +4 c_{4}\tag {3A} \end {align*}
Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = \left (-i \sqrt {3}+2\right )^{3} {\mathrm e}^{\left (-i \sqrt {3}+2\right ) t} c_{1} +3 \left (-i \sqrt {3}+2\right )^{2} {\mathrm e}^{\left (-i \sqrt {3}+2\right ) t} c_{2} +t \left (-i \sqrt {3}+2\right )^{3} {\mathrm e}^{\left (-i \sqrt {3}+2\right ) t} c_{2} +\left (i \sqrt {3}+2\right )^{3} {\mathrm e}^{\left (i \sqrt {3}+2\right ) t} c_{3} +3 \left (i \sqrt {3}+2\right )^{2} {\mathrm e}^{\left (i \sqrt {3}+2\right ) t} c_{4} +t \left (i \sqrt {3}+2\right )^{3} {\mathrm e}^{\left (i \sqrt {3}+2\right ) t} c_{4} \end {align*}
substituting \(y^{\prime \prime \prime } = -1\) and \(t = 0\) in the above gives \begin {align*} -1 = 3 i \left (-3 c_{1} -4 c_{2} +3 c_{3} +4 c_{4} \right ) \sqrt {3}-10 c_{1} +3 c_{2} -10 c_{3} +3 c_{4}\tag {4A} \end {align*}
Equations {1A,2A,3A,4A} are now solved for \(\{c_{1}, c_{2}, c_{3}, c_{4}\}\). Solving for the constants gives \begin {align*} c_{1}&=\frac {\left (2 \sqrt {3}+7 i\right ) \sqrt {3}}{12}\\ c_{2}&=-\frac {\left (7 \sqrt {3}+2 i\right ) \sqrt {3}}{12}\\ c_{3}&=-\frac {\sqrt {3}\, \left (7 i-2 \sqrt {3}\right )}{12}\\ c_{4}&=\frac {\sqrt {3}\, \left (2 i-7 \sqrt {3}\right )}{12} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {7 i \sqrt {3}\, {\mathrm e}^{-\left (-2+i \sqrt {3}\right ) t}}{12}+\frac {{\mathrm e}^{-\left (-2+i \sqrt {3}\right ) t}}{2}-\frac {i \sqrt {3}\, t \,{\mathrm e}^{-\left (-2+i \sqrt {3}\right ) t}}{6}-\frac {7 t \,{\mathrm e}^{-\left (-2+i \sqrt {3}\right ) t}}{4}-\frac {7 i \sqrt {3}\, {\mathrm e}^{\left (i \sqrt {3}+2\right ) t}}{12}+\frac {{\mathrm e}^{\left (i \sqrt {3}+2\right ) t}}{2}+\frac {i \sqrt {3}\, t \,{\mathrm e}^{\left (i \sqrt {3}+2\right ) t}}{6}-\frac {7 t \,{\mathrm e}^{\left (i \sqrt {3}+2\right ) t}}{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{-\left (-2+i \sqrt {3}\right ) t} \left (6+i \left (7-2 t \right ) \sqrt {3}-21 t \right )}{12}+\frac {{\mathrm e}^{\left (i \sqrt {3}+2\right ) t} \left (3+i \left (-\frac {7}{2}+t \right ) \sqrt {3}-\frac {21 t}{2}\right )}{6} \\ \end{align*}
Verification of solutions
\[ y = \frac {{\mathrm e}^{-\left (-2+i \sqrt {3}\right ) t} \left (6+i \left (7-2 t \right ) \sqrt {3}-21 t \right )}{12}+\frac {{\mathrm e}^{\left (i \sqrt {3}+2\right ) t} \left (3+i \left (-\frac {7}{2}+t \right ) \sqrt {3}-\frac {21 t}{2}\right )}{6} \] Verified OK.
Maple trace
`Methods for high order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 36
dsolve([diff(y(t),t$4)-8*diff(y(t),t$3)+30*diff(y(t),t$2)-56*diff(y(t),t)+49*y(t)=0,y(0) = 1, D(y)(0) = 2, (D@@2)(y)(0) = -1, (D@@3)(y)(0) = -1],y(t), singsol=all)
\[ y \left (t \right ) = -\frac {\left (\left (\frac {21 t}{2}-3\right ) \cos \left (\sqrt {3}\, t \right )+\sqrt {3}\, \sin \left (\sqrt {3}\, t \right ) \left (t -\frac {7}{2}\right )\right ) {\mathrm e}^{2 t}}{3} \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 49
DSolve[{y''''[t]-8*y'''[t]+30*y''[t]-56*y'[t]+49*y[t]==0,{y[0]==1,y'[0]==2,y''[0]==-1,y'''[0]==-1}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to -\frac {1}{6} e^{2 t} \left (\sqrt {3} (2 t-7) \sin \left (\sqrt {3} t\right )+3 (7 t-2) \cos \left (\sqrt {3} t\right )\right ) \]