18.30 problem 36

Internal problem ID [14856]
Internal file name [OUTPUT/14536_Monday_April_08_2024_06_26_41_AM_63612407/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Chapter 4 review exercises, page 219
Problem number: 36.
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, _linear, _nonhomogeneous]]

\[ \boxed {y^{\prime \prime \prime \prime }+6 y^{\prime \prime \prime }+18 y^{\prime \prime }+30 y^{\prime }+25 y={\mathrm e}^{-t} \cos \left (2 t \right )+{\mathrm e}^{-2 t} \sin \left (t \right )} \] This is higher order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE And \(y_p\) is a particular solution to the nonhomogeneous ODE. \(y_h\) is the solution to \[ y^{\prime \prime \prime \prime }+6 y^{\prime \prime \prime }+18 y^{\prime \prime }+30 y^{\prime }+25 y = 0 \] The characteristic equation is \[ \lambda ^{4}+6 \lambda ^{3}+18 \lambda ^{2}+30 \lambda +25 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -1-2 i\\ \lambda _2 &= -1+2 i\\ \lambda _3 &= -2-i\\ \lambda _4 &= -2+i \end {align*}

Therefore the homogeneous solution is \[ y_h(t)={\mathrm e}^{\left (-1+2 i\right ) t} c_{1} +{\mathrm e}^{\left (-2+i\right ) t} c_{2} +{\mathrm e}^{\left (-1-2 i\right ) t} c_{3} +{\mathrm e}^{\left (-2-i\right ) t} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{\left (-1+2 i\right ) t} \\ y_2 &= {\mathrm e}^{\left (-2+i\right ) t} \\ y_3 &= {\mathrm e}^{\left (-1-2 i\right ) t} \\ y_4 &= {\mathrm e}^{\left (-2-i\right ) t} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+6 y^{\prime \prime \prime }+18 y^{\prime \prime }+30 y^{\prime }+25 y = {\mathrm e}^{-t} \cos \left (2 t \right )+{\mathrm e}^{-2 t} \sin \left (t \right ) \] Let the particular solution be \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3+U_4 y_4 \] Where \(y_i\) are the basis solutions found above for the homogeneous solution \(y_h\) and \(U_i(t)\) are functions to be determined as follows \[ U_i = (-1)^{n-i} \int { \frac {F(t) W_i(t) }{a W(t)} \, dt} \] Where \(W(t)\) is the Wronskian and \(W_i(t)\) is the Wronskian that results after deleting the last row and the \(i\)-th column of the determinant and \(n\) is the order of the ODE or equivalently, the number of basis solutions, and \(a\) is the coefficient of the leading derivative in the ODE, and \(F(t)\) is the RHS of the ODE. Therefore, the first step is to find the Wronskian \(W \left (t \right )\). This is given by \begin {equation*} W(t) = \begin {vmatrix} y_1&y_2&y_3&y_4\\ y_1'&y_2'&y_3'&y_4'\\ y_1''&y_2''&y_3''&y_4''\\ y_1'''&y_2'''&y_3'''&y_4'''\\ \end {vmatrix} \end {equation*} Substituting the fundamental set of solutions \(y_i\) found above in the Wronskian gives \begin {align*} W &= \left [\begin {array}{cccc} {\mathrm e}^{\left (-1+2 i\right ) t} & {\mathrm e}^{\left (-2+i\right ) t} & {\mathrm e}^{\left (-1-2 i\right ) t} & {\mathrm e}^{\left (-2-i\right ) t} \\ \left (-1+2 i\right ) {\mathrm e}^{\left (-1+2 i\right ) t} & \left (-2+i\right ) {\mathrm e}^{\left (-2+i\right ) t} & \left (-1-2 i\right ) {\mathrm e}^{\left (-1-2 i\right ) t} & \left (-2-i\right ) {\mathrm e}^{\left (-2-i\right ) t} \\ \left (-3-4 i\right ) {\mathrm e}^{\left (-1+2 i\right ) t} & \left (3-4 i\right ) {\mathrm e}^{\left (-2+i\right ) t} & \left (-3+4 i\right ) {\mathrm e}^{\left (-1-2 i\right ) t} & \left (3+4 i\right ) {\mathrm e}^{\left (-2-i\right ) t} \\ \left (11-2 i\right ) {\mathrm e}^{\left (-1+2 i\right ) t} & \left (-2+11 i\right ) {\mathrm e}^{\left (-2+i\right ) t} & \left (11+2 i\right ) {\mathrm e}^{\left (-1-2 i\right ) t} & \left (-2-11 i\right ) {\mathrm e}^{\left (-2-i\right ) t} \end {array}\right ] \\ |W| &= 160 \,{\mathrm e}^{\left (-1+2 i\right ) t} {\mathrm e}^{\left (-2+i\right ) t} {\mathrm e}^{\left (-1-2 i\right ) t} {\mathrm e}^{\left (-2-i\right ) t} \end {align*}

The determinant simplifies to \begin {align*} |W| &= 160 \,{\mathrm e}^{-6 t} \end {align*}

Now we determine \(W_i\) for each \(U_i\). \begin {align*} W_1(t) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{\left (-2+i\right ) t} & {\mathrm e}^{\left (-1-2 i\right ) t} & {\mathrm e}^{\left (-2-i\right ) t} \\ \left (-2+i\right ) {\mathrm e}^{\left (-2+i\right ) t} & \left (-1-2 i\right ) {\mathrm e}^{\left (-1-2 i\right ) t} & \left (-2-i\right ) {\mathrm e}^{\left (-2-i\right ) t} \\ \left (3-4 i\right ) {\mathrm e}^{\left (-2+i\right ) t} & \left (-3+4 i\right ) {\mathrm e}^{\left (-1-2 i\right ) t} & \left (3+4 i\right ) {\mathrm e}^{\left (-2-i\right ) t} \end {array}\right ] \\ &= \left (8-4 i\right ) {\mathrm e}^{\left (-5-2 i\right ) t} \end {align*}

\begin {align*} W_2(t) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{\left (-1+2 i\right ) t} & {\mathrm e}^{\left (-1-2 i\right ) t} & {\mathrm e}^{\left (-2-i\right ) t} \\ \left (-1+2 i\right ) {\mathrm e}^{\left (-1+2 i\right ) t} & \left (-1-2 i\right ) {\mathrm e}^{\left (-1-2 i\right ) t} & \left (-2-i\right ) {\mathrm e}^{\left (-2-i\right ) t} \\ \left (-3-4 i\right ) {\mathrm e}^{\left (-1+2 i\right ) t} & \left (-3+4 i\right ) {\mathrm e}^{\left (-1-2 i\right ) t} & \left (3+4 i\right ) {\mathrm e}^{\left (-2-i\right ) t} \end {array}\right ] \\ &= \left (8-16 i\right ) {\mathrm e}^{\left (-4-i\right ) t} \end {align*}

\begin {align*} W_3(t) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{\left (-1+2 i\right ) t} & {\mathrm e}^{\left (-2+i\right ) t} & {\mathrm e}^{\left (-2-i\right ) t} \\ \left (-1+2 i\right ) {\mathrm e}^{\left (-1+2 i\right ) t} & \left (-2+i\right ) {\mathrm e}^{\left (-2+i\right ) t} & \left (-2-i\right ) {\mathrm e}^{\left (-2-i\right ) t} \\ \left (-3-4 i\right ) {\mathrm e}^{\left (-1+2 i\right ) t} & \left (3-4 i\right ) {\mathrm e}^{\left (-2+i\right ) t} & \left (3+4 i\right ) {\mathrm e}^{\left (-2-i\right ) t} \end {array}\right ] \\ &= \left (8+4 i\right ) {\mathrm e}^{\left (-5+2 i\right ) t} \end {align*}

\begin {align*} W_4(t) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{\left (-1+2 i\right ) t} & {\mathrm e}^{\left (-2+i\right ) t} & {\mathrm e}^{\left (-1-2 i\right ) t} \\ \left (-1+2 i\right ) {\mathrm e}^{\left (-1+2 i\right ) t} & \left (-2+i\right ) {\mathrm e}^{\left (-2+i\right ) t} & \left (-1-2 i\right ) {\mathrm e}^{\left (-1-2 i\right ) t} \\ \left (-3-4 i\right ) {\mathrm e}^{\left (-1+2 i\right ) t} & \left (3-4 i\right ) {\mathrm e}^{\left (-2+i\right ) t} & \left (-3+4 i\right ) {\mathrm e}^{\left (-1-2 i\right ) t} \end {array}\right ] \\ &= \left (8+16 i\right ) {\mathrm e}^{\left (-4+i\right ) t} \end {align*}

Now we are ready to evaluate each \(U_i(t)\). \begin {align*} U_1 &= (-1)^{4-1} \int { \frac {F(t) W_1(t) }{a W(t)} \, dt}\\ &= (-1)^{3} \int { \frac { \left ({\mathrm e}^{-t} \cos \left (2 t \right )+{\mathrm e}^{-2 t} \sin \left (t \right )\right ) \left (\left (8-4 i\right ) {\mathrm e}^{\left (-5-2 i\right ) t}\right )}{\left (1\right ) \left (160 \,{\mathrm e}^{-6 t}\right )} \, dt} \\ &= - \int { \frac {\left (8-4 i\right ) \left ({\mathrm e}^{-t} \cos \left (2 t \right )+{\mathrm e}^{-2 t} \sin \left (t \right )\right ) {\mathrm e}^{\left (-5-2 i\right ) t}}{160 \,{\mathrm e}^{-6 t}} \, dt}\\ &= - \int {\left (\left (\frac {1}{20}-\frac {i}{40}\right ) {\mathrm e}^{-2 i t} \left ({\mathrm e}^{-t} \sin \left (t \right )+\cos \left (2 t \right )\right )\right ) \, dt} \\ &= -\left (\int \left (\frac {1}{20}-\frac {i}{40}\right ) {\mathrm e}^{-2 i t} \left ({\mathrm e}^{-t} \sin \left (t \right )+\cos \left (2 t \right )\right )d t \right ) \end {align*}

\begin {align*} U_2 &= (-1)^{4-2} \int { \frac {F(t) W_2(t) }{a W(t)} \, dt}\\ &= (-1)^{2} \int { \frac { \left ({\mathrm e}^{-t} \cos \left (2 t \right )+{\mathrm e}^{-2 t} \sin \left (t \right )\right ) \left (\left (8-16 i\right ) {\mathrm e}^{\left (-4-i\right ) t}\right )}{\left (1\right ) \left (160 \,{\mathrm e}^{-6 t}\right )} \, dt} \\ &= \int { \frac {\left (8-16 i\right ) \left ({\mathrm e}^{-t} \cos \left (2 t \right )+{\mathrm e}^{-2 t} \sin \left (t \right )\right ) {\mathrm e}^{\left (-4-i\right ) t}}{160 \,{\mathrm e}^{-6 t}} \, dt}\\ &= \int {\left (\left (\frac {1}{20}-\frac {i}{10}\right ) {\mathrm e}^{\left (1-i\right ) t} \left ({\mathrm e}^{-t} \sin \left (t \right )+\cos \left (2 t \right )\right )\right ) \, dt} \\ &= \int \left (\frac {1}{20}-\frac {i}{10}\right ) {\mathrm e}^{\left (1-i\right ) t} \left ({\mathrm e}^{-t} \sin \left (t \right )+\cos \left (2 t \right )\right )d t \end {align*}

\begin {align*} U_3 &= (-1)^{4-3} \int { \frac {F(t) W_3(t) }{a W(t)} \, dt}\\ &= (-1)^{1} \int { \frac { \left ({\mathrm e}^{-t} \cos \left (2 t \right )+{\mathrm e}^{-2 t} \sin \left (t \right )\right ) \left (\left (8+4 i\right ) {\mathrm e}^{\left (-5+2 i\right ) t}\right )}{\left (1\right ) \left (160 \,{\mathrm e}^{-6 t}\right )} \, dt} \\ &= - \int { \frac {\left (8+4 i\right ) \left ({\mathrm e}^{-t} \cos \left (2 t \right )+{\mathrm e}^{-2 t} \sin \left (t \right )\right ) {\mathrm e}^{\left (-5+2 i\right ) t}}{160 \,{\mathrm e}^{-6 t}} \, dt}\\ &= - \int {\left (\left (\frac {1}{20}+\frac {i}{40}\right ) {\mathrm e}^{2 i t} \left ({\mathrm e}^{-t} \sin \left (t \right )+\cos \left (2 t \right )\right )\right ) \, dt} \\ &= -\left (\int \left (\frac {1}{20}+\frac {i}{40}\right ) {\mathrm e}^{2 i t} \left ({\mathrm e}^{-t} \sin \left (t \right )+\cos \left (2 t \right )\right )d t \right ) \end {align*}

\begin {align*} U_4 &= (-1)^{4-4} \int { \frac {F(t) W_4(t) }{a W(t)} \, dt}\\ &= (-1)^{0} \int { \frac { \left ({\mathrm e}^{-t} \cos \left (2 t \right )+{\mathrm e}^{-2 t} \sin \left (t \right )\right ) \left (\left (8+16 i\right ) {\mathrm e}^{\left (-4+i\right ) t}\right )}{\left (1\right ) \left (160 \,{\mathrm e}^{-6 t}\right )} \, dt} \\ &= \int { \frac {\left (8+16 i\right ) \left ({\mathrm e}^{-t} \cos \left (2 t \right )+{\mathrm e}^{-2 t} \sin \left (t \right )\right ) {\mathrm e}^{\left (-4+i\right ) t}}{160 \,{\mathrm e}^{-6 t}} \, dt}\\ &= \int {\left (\left (\frac {1}{20}+\frac {i}{10}\right ) {\mathrm e}^{\left (1+i\right ) t} \left ({\mathrm e}^{-t} \sin \left (t \right )+\cos \left (2 t \right )\right )\right ) \, dt} \\ &= \int \left (\frac {1}{20}+\frac {i}{10}\right ) {\mathrm e}^{\left (1+i\right ) t} \left ({\mathrm e}^{-t} \sin \left (t \right )+\cos \left (2 t \right )\right )d t \end {align*}

Now that all the \(U_i\) functions have been determined, the particular solution is found from \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3+U_4 y_4 \] Hence \begin {equation*} \begin {split} y_p &= \left (-\left (\int \left (\frac {1}{20}-\frac {i}{40}\right ) {\mathrm e}^{-2 i t} \left ({\mathrm e}^{-t} \sin \left (t \right )+\cos \left (2 t \right )\right )d t \right )\right ) \left ({\mathrm e}^{\left (-1+2 i\right ) t}\right ) \\ &+\left (\int \left (\frac {1}{20}-\frac {i}{10}\right ) {\mathrm e}^{\left (1-i\right ) t} \left ({\mathrm e}^{-t} \sin \left (t \right )+\cos \left (2 t \right )\right )d t\right ) \left ({\mathrm e}^{\left (-2+i\right ) t}\right ) \\ &+\left (-\left (\int \left (\frac {1}{20}+\frac {i}{40}\right ) {\mathrm e}^{2 i t} \left ({\mathrm e}^{-t} \sin \left (t \right )+\cos \left (2 t \right )\right )d t \right )\right ) \left ({\mathrm e}^{\left (-1-2 i\right ) t}\right ) \\ &+\left (\int \left (\frac {1}{20}+\frac {i}{10}\right ) {\mathrm e}^{\left (1+i\right ) t} \left ({\mathrm e}^{-t} \sin \left (t \right )+\cos \left (2 t \right )\right )d t\right ) \left ({\mathrm e}^{\left (-2-i\right ) t}\right ) \end {split} \end {equation*} Therefore the particular solution is \[ y_p = -\frac {\left (-\frac {1}{20}+2 \left (-\frac {1}{5}+t \right ) \cos \left (\frac {t}{2}\right )^{4}+\sin \left (\frac {t}{2}\right ) \left (t -\frac {11}{5}\right ) \cos \left (\frac {t}{2}\right )^{3}+\left (\left (t +\frac {1}{5}\right ) {\mathrm e}^{-t}-2 t +\frac {2}{5}\right ) \cos \left (\frac {t}{2}\right )^{2}-\frac {\left (\left (t +\frac {11}{5}\right ) {\mathrm e}^{-t}+t -\frac {11}{5}\right ) \sin \left (\frac {t}{2}\right ) \cos \left (\frac {t}{2}\right )}{2}+\frac {\left (-t -\frac {1}{5}\right ) {\mathrm e}^{-t}}{2}+\frac {t}{4}\right ) {\mathrm e}^{-t}}{5} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{\left (-1+2 i\right ) t} c_{1} +{\mathrm e}^{\left (-2+i\right ) t} c_{2} +{\mathrm e}^{\left (-1-2 i\right ) t} c_{3} +{\mathrm e}^{\left (-2-i\right ) t} c_{4}\right ) + \left (-\frac {\left (-\frac {1}{20}+2 \left (-\frac {1}{5}+t \right ) \cos \left (\frac {t}{2}\right )^{4}+\sin \left (\frac {t}{2}\right ) \left (t -\frac {11}{5}\right ) \cos \left (\frac {t}{2}\right )^{3}+\left (\left (t +\frac {1}{5}\right ) {\mathrm e}^{-t}-2 t +\frac {2}{5}\right ) \cos \left (\frac {t}{2}\right )^{2}-\frac {\left (\left (t +\frac {11}{5}\right ) {\mathrm e}^{-t}+t -\frac {11}{5}\right ) \sin \left (\frac {t}{2}\right ) \cos \left (\frac {t}{2}\right )}{2}+\frac {\left (-t -\frac {1}{5}\right ) {\mathrm e}^{-t}}{2}+\frac {t}{4}\right ) {\mathrm e}^{-t}}{5}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{\left (-1+2 i\right ) t} c_{1} +{\mathrm e}^{\left (-2+i\right ) t} c_{2} +{\mathrm e}^{\left (-1-2 i\right ) t} c_{3} +{\mathrm e}^{\left (-2-i\right ) t} c_{4} -\frac {\left (-\frac {1}{20}+2 \left (-\frac {1}{5}+t \right ) \cos \left (\frac {t}{2}\right )^{4}+\sin \left (\frac {t}{2}\right ) \left (t -\frac {11}{5}\right ) \cos \left (\frac {t}{2}\right )^{3}+\left (\left (t +\frac {1}{5}\right ) {\mathrm e}^{-t}-2 t +\frac {2}{5}\right ) \cos \left (\frac {t}{2}\right )^{2}-\frac {\left (\left (t +\frac {11}{5}\right ) {\mathrm e}^{-t}+t -\frac {11}{5}\right ) \sin \left (\frac {t}{2}\right ) \cos \left (\frac {t}{2}\right )}{2}+\frac {\left (-t -\frac {1}{5}\right ) {\mathrm e}^{-t}}{2}+\frac {t}{4}\right ) {\mathrm e}^{-t}}{5} \\ \end{align*}

Verification of solutions

\[ y = {\mathrm e}^{\left (-1+2 i\right ) t} c_{1} +{\mathrm e}^{\left (-2+i\right ) t} c_{2} +{\mathrm e}^{\left (-1-2 i\right ) t} c_{3} +{\mathrm e}^{\left (-2-i\right ) t} c_{4} -\frac {\left (-\frac {1}{20}+2 \left (-\frac {1}{5}+t \right ) \cos \left (\frac {t}{2}\right )^{4}+\sin \left (\frac {t}{2}\right ) \left (t -\frac {11}{5}\right ) \cos \left (\frac {t}{2}\right )^{3}+\left (\left (t +\frac {1}{5}\right ) {\mathrm e}^{-t}-2 t +\frac {2}{5}\right ) \cos \left (\frac {t}{2}\right )^{2}-\frac {\left (\left (t +\frac {11}{5}\right ) {\mathrm e}^{-t}+t -\frac {11}{5}\right ) \sin \left (\frac {t}{2}\right ) \cos \left (\frac {t}{2}\right )}{2}+\frac {\left (-t -\frac {1}{5}\right ) {\mathrm e}^{-t}}{2}+\frac {t}{4}\right ) {\mathrm e}^{-t}}{5} \] Verified OK.

Maple trace

`Methods for high order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 4; linear nonhomogeneous with symmetry [0,1] 
trying high order linear exact nonhomogeneous 
trying differential order: 4; missing the dependent variable 
checking if the LODE has constant coefficients 
<- constant coefficients successful`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 70

dsolve(diff(y(t),t$4)+6*diff(y(t),t$3)+18*diff(y(t),t$2)+30*diff(y(t),t)+25*y(t)=exp(-t)*cos(2*t)+exp(-2*t)*sin(t),y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {\left (\left (-20 t +400 c_{3} -6\right ) \cos \left (t \right )^{2}-10 \left (t -40 c_{4} -\frac {21}{5}\right ) \sin \left (t \right ) \cos \left (t \right )+10 t -200 c_{3} +3\right ) {\mathrm e}^{-t}}{200}-\frac {{\mathrm e}^{-2 t} \left (\left (t -10 c_{1} +\frac {7}{10}\right ) \cos \left (t \right )-\frac {\left (t +20 c_{2} +\frac {1}{5}\right ) \sin \left (t \right )}{2}\right )}{10} \]

✓ Solution by Mathematica

Time used: 0.438 (sec). Leaf size: 77

DSolve[y''''[t]+6*y'''[t]+18*y''[t]+30*y'[t]+25*y[t]==Exp[-t]*Cos[2*t]+Exp[-2*t]*Sin[t],y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{800} e^{-2 t} \left (-4 (20 t+9-200 c_2) \cos (t)-e^t (40 t-3-800 c_4) \cos (2 t)+8 (5 t+6+100 c_1) \sin (t)-2 e^t (10 t-27-400 c_3) \sin (2 t)\right ) \]