Internal problem ID [12780]
Internal file name [OUTPUT/11433_Friday_November_03_2023_06_32_54_AM_79193795/index.tex]
Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 4. N-th Order Linear Differential Equations. Exercises 4.5, page 221
Problem number: 1.
ODE order: 3.
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
[[_3rd_order, _missing_x]]
\[ \boxed {y^{\prime \prime \prime }-3 y^{\prime \prime }-4 y^{\prime }+12 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 5, y^{\prime \prime }\left (0\right ) = -1] \end {align*}
The characteristic equation is \[ \lambda ^{3}-3 \lambda ^{2}-4 \lambda +12 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= 3\\ \lambda _3 &= -2 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{2 x}+c_{3} {\mathrm e}^{3 x} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{-2 x}\\ y_2 &= {\mathrm e}^{2 x}\\ y_3 &= {\mathrm e}^{3 x} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{2 x}+c_{3} {\mathrm e}^{3 x} \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 \(x = 0\) in the above gives \begin {align*} 1 = c_{1} +c_{2} +c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -2 c_{1} {\mathrm e}^{-2 x}+2 c_{2} {\mathrm e}^{2 x}+3 c_{3} {\mathrm e}^{3 x} \end {align*}
substituting \(y^{\prime } = 5\) and \(x = 0\) in the above gives \begin {align*} 5 = -2 c_{1} +2 c_{2} +3 c_{3}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = 4 c_{1} {\mathrm e}^{-2 x}+4 c_{2} {\mathrm e}^{2 x}+9 c_{3} {\mathrm e}^{3 x} \end {align*}
substituting \(y^{\prime \prime } = -1\) and \(x = 0\) in the above gives \begin {align*} -1 = 4 c_{1} +4 c_{2} +9 c_{3}\tag {3A} \end {align*}
Equations {1A,2A,3A} are now solved for \(\{c_{1}, c_{2}, c_{3}\}\). Solving for the constants gives \begin {align*} c_{1}&=-1\\ c_{2}&=3\\ c_{3}&=-1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -{\mathrm e}^{-2 x}+3 \,{\mathrm e}^{2 x}-{\mathrm e}^{3 x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -{\mathrm e}^{-2 x}+3 \,{\mathrm e}^{2 x}-{\mathrm e}^{3 x} \\ \end{align*}
Verification of solutions
\[ y = -{\mathrm e}^{-2 x}+3 \,{\mathrm e}^{2 x}-{\mathrm e}^{3 x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }-3 y^{\prime \prime }-4 y^{\prime }+12 y=0, y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=5, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (x \right ) \\ {} & {} & y_{1}\left (x \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=3 y_{3}\left (x \right )+4 y_{2}\left (x \right )-12 y_{1}\left (x \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=y_{1}^{\prime }\left (x \right ), y_{3}\left (x \right )=y_{2}^{\prime }\left (x \right ), y_{3}^{\prime }\left (x \right )=3 y_{3}\left (x \right )+4 y_{2}\left (x \right )-12 y_{1}\left (x \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{c} y_{1}\left (x \right ) \\ y_{2}\left (x \right ) \\ y_{3}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -12 & 4 & 3 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -12 & 4 & 3 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=A \cdot {\moverset {\rightarrow }{y}}\left (x \right ) \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [-2, \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-2, \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{3 x}\cdot \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {General solution to the system of ODEs}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]+c_{3} {\mathrm e}^{3 x}\cdot \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (4 c_{3} {\mathrm e}^{5 x}+9 c_{2} {\mathrm e}^{4 x}+9 c_{1} \right ) {\mathrm e}^{-2 x}}{36} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=\frac {c_{3}}{9}+\frac {c_{2}}{4}+\frac {c_{1}}{4} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\left (20 c_{3} {\mathrm e}^{5 x}+36 c_{2} {\mathrm e}^{4 x}\right ) {\mathrm e}^{-2 x}}{36}-\frac {\left (4 c_{3} {\mathrm e}^{5 x}+9 c_{2} {\mathrm e}^{4 x}+9 c_{1} \right ) {\mathrm e}^{-2 x}}{18} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=5 \\ {} & {} & 5=\frac {c_{3}}{3}+\frac {c_{2}}{2}-\frac {c_{1}}{2} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\left (100 c_{3} {\mathrm e}^{5 x}+144 c_{2} {\mathrm e}^{4 x}\right ) {\mathrm e}^{-2 x}}{36}-\frac {\left (20 c_{3} {\mathrm e}^{5 x}+36 c_{2} {\mathrm e}^{4 x}\right ) {\mathrm e}^{-2 x}}{9}+\frac {\left (4 c_{3} {\mathrm e}^{5 x}+9 c_{2} {\mathrm e}^{4 x}+9 c_{1} \right ) {\mathrm e}^{-2 x}}{9} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-1 \\ {} & {} & -1=c_{1} +c_{2} +c_{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =-4, c_{2} =12, c_{3} =-9\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (-{\mathrm e}^{5 x}+3 \,{\mathrm e}^{4 x}-1\right ) {\mathrm e}^{-2 x} \end {array} \]
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 23
dsolve([diff(y(x),x$3)-3*diff(y(x),x$2)-4*diff(y(x),x)+12*y(x)=0,y(0) = 1, D(y)(0) = 5, (D@@2)(y)(0) = -1],y(x), singsol=all)
\[ y \left (x \right ) = \left (-{\mathrm e}^{5 x}+3 \,{\mathrm e}^{4 x}-1\right ) {\mathrm e}^{-2 x} \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 26
DSolve[{y'''[x]-3*y''[x]-4*y'[x]+12*y[x]==0,{y[0]==1,y'[0]==5,y''[0]==-1}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -e^{-2 x} \left (-3 e^{4 x}+e^{5 x}+1\right ) \]