Internal problem ID [297]
Internal file name [OUTPUT/297_Sunday_June_05_2022_01_38_22_AM_69806262/index.tex]
Book: Differential equations and linear algebra, 4th ed., Edwards and Penney
Section: Section 5.3, Higher-Order Linear Differential Equations. Homogeneous Equations with
Constant Coefficients. Page 300
Problem number: problem 25.
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]]
\[ \boxed {3 y^{\prime \prime \prime }+2 y^{\prime \prime }=0} \] With initial conditions \begin {align*} [y \left (0\right ) = -1, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 1] \end {align*}
The characteristic equation is \[ 3 \lambda ^{3}+2 \lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -{\frac {2}{3}}\\ \lambda _2 &= 0\\ \lambda _3 &= 0 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{2} x +c_{1} +{\mathrm e}^{-\frac {2 x}{3}} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= 1\\ y_2 &= x\\ y_3 &= {\mathrm e}^{-\frac {2 x}{3}} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{2} x +c_{1} +{\mathrm e}^{-\frac {2 x}{3}} c_{3} \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_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{2} -\frac {2 \,{\mathrm e}^{-\frac {2 x}{3}} c_{3}}{3} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = c_{2} -\frac {2 c_{3}}{3}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = \frac {4 \,{\mathrm e}^{-\frac {2 x}{3}} c_{3}}{9} \end {align*}
substituting \(y^{\prime \prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = \frac {4 c_{3}}{9}\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}&=-{\frac {13}{4}}\\ c_{2}&={\frac {3}{2}}\\ c_{3}&={\frac {9}{4}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -\frac {13}{4}+\frac {3 x}{2}+\frac {9 \,{\mathrm e}^{-\frac {2 x}{3}}}{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {13}{4}+\frac {3 x}{2}+\frac {9 \,{\mathrm e}^{-\frac {2 x}{3}}}{4} \\ \end{align*}
Verification of solutions
\[ y = -\frac {13}{4}+\frac {3 x}{2}+\frac {9 \,{\mathrm e}^{-\frac {2 x}{3}}}{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [3 y^{\prime \prime \prime }+2 y^{\prime \prime }=0, y \left (0\right )=-1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0, 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 } \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=-\frac {2 y^{\prime \prime }}{3} \\ \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 \prime }+\frac {2 y^{\prime \prime }}{3}=0 \\ \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 )=-\frac {2 y_{3}\left (x \right )}{3} \\ & {} & \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 )=-\frac {2 y_{3}\left (x \right )}{3}\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 \\ 0 & 0 & -\frac {2}{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 \\ 0 & 0 & -\frac {2}{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 [-\frac {2}{3}, \left [\begin {array}{c} \frac {9}{4} \\ -\frac {3}{2} \\ 1 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-\frac {2}{3}, \left [\begin {array}{c} \frac {9}{4} \\ -\frac {3}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-\frac {2 x}{3}}\cdot \left [\begin {array}{c} \frac {9}{4} \\ -\frac {3}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}=\left [\begin {array}{c} 0 \\ 0 \\ 0 \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}^{-\frac {2 x}{3}}\cdot \left [\begin {array}{c} \frac {9}{4} \\ -\frac {3}{2} \\ 1 \end {array}\right ]+\left [\begin {array}{c} c_{2} \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {9 c_{1} {\mathrm e}^{-\frac {2 x}{3}}}{4}+c_{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=-1 \\ {} & {} & -1=\frac {9 c_{1}}{4}+c_{2} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {3 c_{1} {\mathrm e}^{-\frac {2 x}{3}}}{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-\frac {3 c_{1}}{2} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=c_{1} {\mathrm e}^{-\frac {2 x}{3}} \\ \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} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ \bullet & {} & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \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: 0.016 (sec). Leaf size: 15
dsolve([3*diff(y(x),x$3)+2*diff(y(x),x$2)=0,y(0) = -1, D(y)(0) = 0, (D@@2)(y)(0) = 1],y(x), singsol=all)
\[ y \left (x \right ) = -\frac {13}{4}+\frac {3 x}{2}+\frac {9 \,{\mathrm e}^{-\frac {2 x}{3}}}{4} \]
✓ Solution by Mathematica
Time used: 0.038 (sec). Leaf size: 23
DSolve[{3*y'''[x]+2*y''[x]==0,{y[0]==1,y'[0]==-1,y''[0]==3}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{4} \left (14 x+27 e^{-2 x/3}-23\right ) \]