Internal problem ID [12789]
Internal file name [OUTPUT/11442_Saturday_November_04_2023_08_47_20_AM_55985488/index.tex]
Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 5. The Laplace Transform Method. Exercises 5.2, page 248
Problem number: 6.
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]]
\[ \boxed {y^{\prime \prime }+y^{\prime }-2 y=x \,{\mathrm e}^{x}-3 x^{2}} \] 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 )+s Y \left (s \right )-y \left (0\right )-2 Y \left (s \right ) = \frac {1}{\left (s -1\right )^{2}}-\frac {6}{s^{3}}\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} +s Y \left (s \right )-c_{1} -2 Y \left (s \right ) = \frac {1}{\left (s -1\right )^{2}}-\frac {6}{s^{3}} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {c_{1} s^{6}-c_{1} s^{5}+c_{2} s^{5}-c_{1} s^{4}-2 c_{2} s^{4}+c_{1} s^{3}+c_{2} s^{3}+s^{3}-6 s^{2}+12 s -6}{\left (s -1\right )^{2} s^{3} \left (s^{2}+s -2\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= -\frac {1}{9 \left (s -1\right )^{2}}+\frac {1}{3 \left (s -1\right )^{3}}+\frac {-\frac {53}{27}+\frac {2 c_{1}}{3}+\frac {c_{2}}{3}}{s -1}+\frac {\frac {c_{1}}{3}-\frac {c_{2}}{3}-\frac {31}{108}}{s +2}+\frac {3}{s^{3}}+\frac {3}{2 s^{2}}+\frac {9}{4 s} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {1}{9 \left (s -1\right )^{2}}\right ) &= -\frac {x \,{\mathrm e}^{x}}{9}\\ \mathcal {L}^{-1}\left (\frac {1}{3 \left (s -1\right )^{3}}\right ) &= \frac {{\mathrm e}^{x} x^{2}}{6}\\ \mathcal {L}^{-1}\left (\frac {-\frac {53}{27}+\frac {2 c_{1}}{3}+\frac {c_{2}}{3}}{s -1}\right ) &= \frac {\left (-53+18 c_{1} +9 c_{2} \right ) {\mathrm e}^{x}}{27}\\ \mathcal {L}^{-1}\left (\frac {\frac {c_{1}}{3}-\frac {c_{2}}{3}-\frac {31}{108}}{s +2}\right ) &= \frac {\left (36 c_{1} -36 c_{2} -31\right ) {\mathrm e}^{-2 x}}{108}\\ \mathcal {L}^{-1}\left (\frac {3}{s^{3}}\right ) &= \frac {3 x^{2}}{2}\\ \mathcal {L}^{-1}\left (\frac {3}{2 s^{2}}\right ) &= \frac {3 x}{2}\\ \mathcal {L}^{-1}\left (\frac {9}{4 s}\right ) &= {\frac {9}{4}} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {9}{4}+\frac {3 x^{2}}{2}+\frac {3 x}{2}+\frac {{\mathrm e}^{x} \left (9 x^{2}+36 c_{1} +18 c_{2} -6 x -106\right )}{54}+\frac {\left (36 c_{1} -36 c_{2} -31\right ) {\mathrm e}^{-2 x}}{108} \] Simplifying the solution gives \[ y = \frac {3 \left (-\frac {31}{162}+\frac {\left (x^{2}-\frac {2}{3} x +4 c_{1} +2 c_{2} -\frac {106}{9}\right ) {\mathrm e}^{3 x}}{9}+\left (x^{2}+x +\frac {3}{2}\right ) {\mathrm e}^{2 x}+\frac {2 c_{1}}{9}-\frac {2 c_{2}}{9}\right ) {\mathrm e}^{-2 x}}{2} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {3 \left (-\frac {31}{162}+\frac {\left (x^{2}-\frac {2}{3} x +4 c_{1} +2 c_{2} -\frac {106}{9}\right ) {\mathrm e}^{3 x}}{9}+\left (x^{2}+x +\frac {3}{2}\right ) {\mathrm e}^{2 x}+\frac {2 c_{1}}{9}-\frac {2 c_{2}}{9}\right ) {\mathrm e}^{-2 x}}{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {3 \left (-\frac {31}{162}+\frac {\left (x^{2}-\frac {2}{3} x +4 c_{1} +2 c_{2} -\frac {106}{9}\right ) {\mathrm e}^{3 x}}{9}+\left (x^{2}+x +\frac {3}{2}\right ) {\mathrm e}^{2 x}+\frac {2 c_{1}}{9}-\frac {2 c_{2}}{9}\right ) {\mathrm e}^{-2 x}}{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+y^{\prime }-2 y=x \,{\mathrm e}^{x}-3 x^{2} \\ \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}+r -2=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +2\right ) \left (r -1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2, 1\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )={\mathrm e}^{-2 x} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )={\mathrm e}^{x} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right )+y_{p}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{x}+y_{p}\left (x \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (x \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 (x \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (x \right )=-y_{1}\left (x \right ) \left (\int \frac {y_{2}\left (x \right ) f \left (x \right )}{W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )}d x \right )+y_{2}\left (x \right ) \left (\int \frac {y_{1}\left (x \right ) f \left (x \right )}{W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )}d x \right ), f \left (x \right )=x \,{\mathrm e}^{x}-3 x^{2}\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=\left [\begin {array}{cc} {\mathrm e}^{-2 x} & {\mathrm e}^{x} \\ -2 \,{\mathrm e}^{-2 x} & {\mathrm e}^{x} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=3 \,{\mathrm e}^{-x} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & y_{p}\left (x \right )=-\frac {\left ({\mathrm e}^{3 x} \left (\int \left (3 \,{\mathrm e}^{-x} x^{2}-x \right )d x \right )+\int {\mathrm e}^{2 x} x \left ({\mathrm e}^{x}-3 x \right )d x \right ) {\mathrm e}^{-2 x}}{3} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=\frac {9}{4}+\frac {\left (9 x^{2}-6 x +2\right ) {\mathrm e}^{x}}{54}+\frac {3 x^{2}}{2}+\frac {3 x}{2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{x}+\frac {9}{4}+\frac {\left (9 x^{2}-6 x +2\right ) {\mathrm e}^{x}}{54}+\frac {3 x^{2}}{2}+\frac {3 x}{2} \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: 5.812 (sec). Leaf size: 52
dsolve(diff(y(x),x$2)+diff(y(x),x)-2*y(x)=x*exp(x)-3*x^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {9}{4}+\frac {3 x}{2}+\frac {3 x^{2}}{2}+\frac {{\mathrm e}^{x} \left (9 x^{2}+18 D\left (y \right )\left (0\right )+36 y \left (0\right )-6 x -106\right )}{54}+\frac {\left (36 y \left (0\right )-36 D\left (y \right )\left (0\right )-31\right ) {\mathrm e}^{-2 x}}{108} \]
✓ Solution by Mathematica
Time used: 0.313 (sec). Leaf size: 49
DSolve[y''[x]+y'[x]-2*y[x]==x*Exp[x]-3*x^2,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {3}{4} \left (2 x^2+2 x+3\right )+\frac {1}{54} e^x \left (9 x^2-6 x+2+54 c_2\right )+c_1 e^{-2 x} \]