Internal problem ID [12790]
Internal file name [OUTPUT/11443_Saturday_November_04_2023_08_47_21_AM_22023351/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: 7.
ODE order: 4.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_laplace"
Maple gives the following as the ode type
[[_high_order, _missing_y]]
\[ \boxed {y^{\prime \prime \prime \prime }-2 y^{\prime \prime \prime }+y^{\prime \prime }=x \,{\mathrm e}^{x}-3 x^{2}} \] Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] 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 )\\ \mathcal {L}\left (y^{\prime \prime \prime }\right ) &= s^3 Y(s) - y''(0) - s y'(0) - s^2 y \left (0\right )\\ \mathcal {L}\left (y^{\prime \prime \prime \prime }\right ) &= s^4 Y(s) - y'''(0) - s y''(0) - s^2 y'(0)- s^3 y \left (0\right ) \end {align*}
The given ode becomes an algebraic equation in the Laplace domain \[ s^{4} Y \left (s \right )-y^{\prime \prime \prime }\left (0\right )-s y^{\prime \prime }\left (0\right )-s^{2} y^{\prime }\left (0\right )-s^{3} y \left (0\right )-2 s^{3} Y \left (s \right )+2 y^{\prime \prime }\left (0\right )+2 s y^{\prime }\left (0\right )+2 s^{2} y \left (0\right )+s^{2} Y \left (s \right )-y^{\prime }\left (0\right )-s y \left (0\right ) = \frac {1}{\left (s -1\right )^{2}}-\frac {6}{s^{3}}\tag {1} \] But the initial conditions are \begin {align*} y \left (0\right )&=c_{1}\\ y^{\prime }\left (0\right )&=c_{2}\\ y^{\prime \prime }\left (0\right )&=c_{3}\\ y^{\prime \prime \prime }\left (0\right )&=c_{4} \end {align*}
Substituting these initial conditions in above in Eq (1) gives \[ s^{4} Y \left (s \right )-c_{4} -s c_{3} -s^{2} c_{2} -s^{3} c_{1} -2 s^{3} Y \left (s \right )+2 c_{3} +2 s c_{2} +2 s^{2} c_{1} +s^{2} Y \left (s \right )-c_{2} -s c_{1} = \frac {1}{\left (s -1\right )^{2}}-\frac {6}{s^{3}} \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = \frac {c_{1} s^{8}-4 c_{1} s^{7}+c_{2} s^{7}+6 c_{1} s^{6}-4 c_{2} s^{6}+c_{3} s^{6}-4 c_{1} s^{5}+6 c_{2} s^{5}-4 c_{3} s^{5}+c_{4} s^{5}+c_{1} s^{4}-4 c_{2} s^{4}+5 c_{3} s^{4}-2 c_{4} s^{4}+c_{2} s^{3}-2 c_{3} s^{3}+c_{4} s^{3}+s^{3}-6 s^{2}+12 s -6}{\left (s -1\right )^{2} s^{5} \left (s^{2}-2 s +1\right )} \] Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{\left (s -1\right )^{4}}-\frac {6}{s^{5}}+\frac {-c_{3} +c_{4} -3}{\left (s -1\right )^{2}}+\frac {-26-3 c_{3} +2 c_{4} +c_{1}}{s}-\frac {12}{s^{4}}+\frac {3 c_{3} -2 c_{4} +26}{s -1}+\frac {-23+c_{2} -2 c_{3} +c_{4}}{s^{2}}-\frac {18}{s^{3}}-\frac {2}{\left (s -1\right )^{3}} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{\left (s -1\right )^{4}}\right ) &= \frac {x^{3} {\mathrm e}^{x}}{6}\\ \mathcal {L}^{-1}\left (-\frac {6}{s^{5}}\right ) &= -\frac {x^{4}}{4}\\ \mathcal {L}^{-1}\left (\frac {-c_{3} +c_{4} -3}{\left (s -1\right )^{2}}\right ) &= \left (-c_{3} +c_{4} -3\right ) x \,{\mathrm e}^{x}\\ \mathcal {L}^{-1}\left (\frac {-26-3 c_{3} +2 c_{4} +c_{1}}{s}\right ) &= -26-3 c_{3} +2 c_{4} +c_{1}\\ \mathcal {L}^{-1}\left (-\frac {12}{s^{4}}\right ) &= -2 x^{3}\\ \mathcal {L}^{-1}\left (\frac {3 c_{3} -2 c_{4} +26}{s -1}\right ) &= \left (3 c_{3} -2 c_{4} +26\right ) {\mathrm e}^{x}\\ \mathcal {L}^{-1}\left (\frac {-23+c_{2} -2 c_{3} +c_{4}}{s^{2}}\right ) &= \left (-23+c_{2} -2 c_{3} +c_{4} \right ) x\\ \mathcal {L}^{-1}\left (-\frac {18}{s^{3}}\right ) &= -9 x^{2}\\ \mathcal {L}^{-1}\left (-\frac {2}{\left (s -1\right )^{3}}\right ) &= -{\mathrm e}^{x} x^{2} \end {align*}
Adding the above results and simplifying gives \[ y=-26-3 c_{3} +2 c_{4} +c_{1} -9 x^{2}-\frac {x^{4}}{4}-2 x^{3}+\frac {{\mathrm e}^{x} \left (x^{3}-6 c_{3} x +6 c_{4} x -6 x^{2}+18 c_{3} -12 c_{4} -18 x +156\right )}{6}+\left (-23+c_{2} -2 c_{3} +c_{4} \right ) x \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -26-3 c_{3} +2 c_{4} +c_{1} -9 x^{2}-\frac {x^{4}}{4}-2 x^{3}+\frac {{\mathrm e}^{x} \left (x^{3}-6 c_{3} x +6 c_{4} x -6 x^{2}+18 c_{3} -12 c_{4} -18 x +156\right )}{6}+\left (-23+c_{2} -2 c_{3} +c_{4} \right ) x \\ \end{align*}
Verification of solutions
\[ y = -26-3 c_{3} +2 c_{4} +c_{1} -9 x^{2}-\frac {x^{4}}{4}-2 x^{3}+\frac {{\mathrm e}^{x} \left (x^{3}-6 c_{3} x +6 c_{4} x -6 x^{2}+18 c_{3} -12 c_{4} -18 x +156\right )}{6}+\left (-23+c_{2} -2 c_{3} +c_{4} \right ) x \] 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] -> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = _a*exp(_a)-3*_a^2+2*(diff(_b(_a), _a))-_b(_a), _b(_a)` *** Sublevel 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 <- differential order: 4; linear nonhomogeneous with symmetry [0,1] successful`
✓ Solution by Maple
Time used: 5.985 (sec). Leaf size: 79
dsolve(diff(y(x),x$4)-2*diff(y(x),x$3)+diff(y(x),x$2)=x*exp(x)-3*x^2,y(x), singsol=all)
\[ y \left (x \right ) = -26-\frac {x^{4}}{4}-9 x^{2}-2 x^{3}+y \left (0\right )+\frac {{\mathrm e}^{x} \left (x^{3}+6 x D^{\left (3\right )}\left (y \right )\left (0\right )-6 x D^{\left (2\right )}\left (y \right )\left (0\right )-6 x^{2}-12 D^{\left (3\right )}\left (y \right )\left (0\right )+18 D^{\left (2\right )}\left (y \right )\left (0\right )-18 x +156\right )}{6}-D^{\left (2\right )}\left (y \right )\left (0\right ) \left (3+2 x \right )+D^{\left (3\right )}\left (y \right )\left (0\right ) \left (x +2\right )+x \left (-23+D\left (y \right )\left (0\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.812 (sec). Leaf size: 59
DSolve[y''''[x]-2*y'''[x]+y''[x]==x*Exp[x]-3*x^2,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {x^4}{4}-2 x^3-9 x^2+e^x \left (\frac {x^3}{6}-x^2+(3+c_2) x-4+c_1-2 c_2\right )+c_4 x+c_3 \]