Internal problem ID [15277]
Internal file name [OUTPUT/15278_Wednesday_May_08_2024_03_54_41_PM_87940267/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 15.3 Nonhomogeneous linear equations with
constant coefficients. Trial and error method. Exercises page 132
Problem number: 508.
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, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime \prime \prime }-4 y^{\prime \prime \prime }+6 y^{\prime \prime }-4 y^{\prime }+y={\mathrm e}^{x}} \] 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 \[ y^{\prime \prime \prime \prime }-4 y^{\prime \prime \prime }+6 y^{\prime \prime }-4 y^{\prime }+y = 0 \] The characteristic equation is \[ \lambda ^{4}-4 \lambda ^{3}+6 \lambda ^{2}-4 \lambda +1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 1\\ \lambda _3 &= 1\\ \lambda _4 &= 1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{x}+x \,{\mathrm e}^{x} c_{2} +x^{2} {\mathrm e}^{x} c_{3} +x^{3} {\mathrm e}^{x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{x} \\ y_2 &= {\mathrm e}^{x} x \\ y_3 &= {\mathrm e}^{x} x^{2} \\ y_4 &= x^{3} {\mathrm e}^{x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-4 y^{\prime \prime \prime }+6 y^{\prime \prime }-4 y^{\prime }+y = {\mathrm e}^{x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x^{3} {\mathrm e}^{x}, {\mathrm e}^{x} x, {\mathrm e}^{x} x^{2}, {\mathrm e}^{x}\} \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{x} x\}] \] Since \({\mathrm e}^{x} x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{x} x^{2}\}] \] Since \({\mathrm e}^{x} x^{2}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{3} {\mathrm e}^{x}\}] \] Since \(x^{3} {\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{4} {\mathrm e}^{x}\}] \] Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set. \[ y_p = A_{1} x^{4} {\mathrm e}^{x} \] The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives \[ 24 A_{1} {\mathrm e}^{x} = {\mathrm e}^{x} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{24}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x^{4} {\mathrm e}^{x}}{24} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{x}+x \,{\mathrm e}^{x} c_{2} +x^{2} {\mathrm e}^{x} c_{3} +x^{3} {\mathrm e}^{x} c_{4}\right ) + \left (\frac {x^{4} {\mathrm e}^{x}}{24}\right ) \\ \end{align*} Which simplifies to \[ y = {\mathrm e}^{x} \left (c_{4} x^{3}+c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {x^{4} {\mathrm e}^{x}}{24} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{x} \left (c_{4} x^{3}+c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {x^{4} {\mathrm e}^{x}}{24} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{x} \left (c_{4} x^{3}+c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {x^{4} {\mathrm e}^{x}}{24} \] 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: 27
dsolve(diff(y(x),x$4)-4*diff(y(x),x$3)+6*diff(y(x),x$2)-4*diff(y(x),x)+y(x)=exp(x),y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{x} \left (\frac {1}{24} x^{4}+c_{1} +c_{2} x +c_{3} x^{2}+c_{4} x^{3}\right ) \]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 39
DSolve[y''''[x]-4*y'''[x]+6*y''[x]-4*y'[x]+y[x]==Exp[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{24} e^x \left (x^4+24 c_4 x^3+24 c_3 x^2+24 c_2 x+24 c_1\right ) \]