Internal problem ID [1582]
Internal file name [OUTPUT/1583_Sunday_June_05_2022_02_23_33_AM_87938238/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 9 Introduction to Linear Higher Order Equations. Section 9.4. Variation of
Parameters for Higher Order Equations. Page 503
Problem number: section 9.4, problem 30.
ODE order: 4.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_ODE_non_constant_coefficients_of_type_Euler"
Maple gives the following as the ode type
[[_high_order, _exact, _linear, _nonhomogeneous]]
\[ \boxed {x^{4} y^{\prime \prime \prime \prime }+3 x^{3} y^{\prime \prime \prime }-x^{2} y^{\prime \prime }+2 y^{\prime } x -2 y=9 x^{2}} \] With initial conditions \begin {align*} [y \left (1\right ) = -7, y^{\prime }\left (1\right ) = -11, y^{\prime \prime }\left (1\right ) = -5, y^{\prime \prime \prime }\left (1\right ) = 6] \end {align*}
This is higher order nonhomogeneous Euler type ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous Euler ODE And \(y_p\) is a particular solution to the nonhomogeneous Euler ODE. \(y_h\) is the solution to \[ x^{4} y^{\prime \prime \prime \prime }+3 x^{3} y^{\prime \prime \prime }-x^{2} y^{\prime \prime }+2 y^{\prime } x -2 y = 0 \] This is Euler ODE of higher order. Let \(y = x^{\lambda }\). Hence \begin {align*} y^{\prime } &= \lambda \,x^{\lambda -1}\\ y^{\prime \prime } &= \lambda \left (\lambda -1\right ) x^{\lambda -2}\\ y^{\prime \prime \prime } &= \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}\\ y^{\prime \prime \prime \prime } &= \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) x^{\lambda -4} \end {align*}
Substituting these back into \[ x^{4} y^{\prime \prime \prime \prime }+3 x^{3} y^{\prime \prime \prime }-x^{2} y^{\prime \prime }+2 y^{\prime } x -2 y = 9 x^{2} \] gives \[ 2 x \lambda \,x^{\lambda -1}-x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+3 x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}+x^{4} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) x^{\lambda -4}-2 x^{\lambda } = 0 \] Which simplifies to \[ 2 \lambda \,x^{\lambda }-\lambda \left (\lambda -1\right ) x^{\lambda }+3 \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) x^{\lambda }-2 x^{\lambda } = 0 \] And since \(x^{\lambda }\neq 0\) then dividing through by \(x^{\lambda }\), the above becomes
\[ 2 \lambda -\lambda \left (\lambda -1\right )+3 \lambda \left (\lambda -1\right ) \left (\lambda -2\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right )-2 = 0 \] Simplifying gives the characteristic equation as \[ \left (\lambda +1\right ) \left (\lambda -2\right ) \left (\lambda -1\right )^{2} = 0 \] Solving the above gives the following roots \begin {align*} \lambda _1 &= -1\\ \lambda _2 &= 2\\ \lambda _3 &= 1\\ \lambda _4 &= 1 \end {align*}
This table summarises the result
| root | multiplicity | type of root |
| \(-1\) | \(1\) | real root |
| \(1\) | \(2\) | real root |
| \(2\) | \(1\) | real root |
The solution is generated by going over the above table. For each real root \(\lambda \) of multiplicity one generates a \(c_1x^{\lambda }\) basis solution. Each real root of multiplicty two, generates \(c_1x^{\lambda }\) and \(c_2x^{\lambda } \ln \left (x \right )\) basis solutions. Each real root of multiplicty three, generates \(c_1x^{\lambda }\) and \(c_2x^{\lambda } \ln \left (x \right )\) and \(c_3x^{\lambda } \ln \left (x \right )^{2}\) basis solutions, and so on. Each complex root \(\alpha \pm i \beta \) of multiplicity one generates \(x^{\alpha } \left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And each complex root \(\alpha \pm i \beta \) of multiplicity two generates \(\ln \left (x \right ) x^{\alpha }\left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And each complex root \(\alpha \pm i \beta \) of multiplicity three generates \(\ln \left (x \right )^{2} x^{\alpha }\left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And so on. Using the above show that the solution is
\[ y = \frac {c_{1}}{x}+c_{2} x +\ln \left (x \right ) c_{3} x +c_{4} x^{2} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= \frac {1}{x} \\ y_2 &= x \\ y_3 &= x \ln \left (x \right ) \\ y_4 &= x^{2} \\ \end{align*} Now the particular solution to the given ODE is found \[ x^{4} y^{\prime \prime \prime \prime }+3 x^{3} y^{\prime \prime \prime }-x^{2} y^{\prime \prime }+2 y^{\prime } x -2 y = 9 x^{2} \] Let the particular solution be \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3+U_4 y_4 \] Where \(y_i\) are the basis solutions found above for the homogeneous solution \(y_h\) and \(U_i(x)\) are functions to be determined as follows \[ U_i = (-1)^{n-i} \int { \frac {F(x) W_i(x) }{a W(x)} \, dx} \] Where \(W(x)\) is the Wronskian and \(W_i(x)\) is the Wronskian that results after deleting the last row and the \(i\)-th column of the determinant and \(n\) is the order of the ODE or equivalently, the number of basis solutions, and \(a\) is the coefficient of the leading derivative in the ODE, and \(F(x)\) is the RHS of the ODE. Therefore, the first step is to find the Wronskian \(W \left (x \right )\). This is given by \begin {equation*} W(x) = \begin {vmatrix} y_1&y_2&y_3&y_4\\ y_1'&y_2'&y_3'&y_4'\\ y_1''&y_2''&y_3''&y_4''\\ y_1'''&y_2'''&y_3'''&y_4'''\\ \end {vmatrix} \end {equation*} Substituting the fundamental set of solutions \(y_i\) found above in the Wronskian gives \begin {align*} W &= \left [\begin {array}{cccc} \frac {1}{x} & x & x \ln \left (x \right ) & x^{2} \\ -\frac {1}{x^{2}} & 1 & \ln \left (x \right )+1 & 2 x \\ \frac {2}{x^{3}} & 0 & \frac {1}{x} & 2 \\ -\frac {6}{x^{4}} & 0 & -\frac {1}{x^{2}} & 0 \end {array}\right ] \\ |W| &= \frac {12}{x^{3}} \end {align*}
The determinant simplifies to \begin {align*} |W| &= \frac {12}{x^{3}} \end {align*}
Now we determine \(W_i\) for each \(U_i\). \begin {align*} W_1(x) &= \det \,\left [\begin {array}{ccc} x & x \ln \left (x \right ) & x^{2} \\ 1 & \ln \left (x \right )+1 & 2 x \\ 0 & \frac {1}{x} & 2 \end {array}\right ] \\ &= x \end {align*}
\begin {align*} W_2(x) &= \det \,\left [\begin {array}{ccc} \frac {1}{x} & x \ln \left (x \right ) & x^{2} \\ -\frac {1}{x^{2}} & \ln \left (x \right )+1 & 2 x \\ \frac {2}{x^{3}} & \frac {1}{x} & 2 \end {array}\right ] \\ &= \frac {6 \ln \left (x \right )-3}{x} \end {align*}
\begin {align*} W_3(x) &= \det \,\left [\begin {array}{ccc} \frac {1}{x} & x & x^{2} \\ -\frac {1}{x^{2}} & 1 & 2 x \\ \frac {2}{x^{3}} & 0 & 2 \end {array}\right ] \\ &= \frac {6}{x} \end {align*}
\begin {align*} W_4(x) &= \det \,\left [\begin {array}{ccc} \frac {1}{x} & x & x \ln \left (x \right ) \\ -\frac {1}{x^{2}} & 1 & \ln \left (x \right )+1 \\ \frac {2}{x^{3}} & 0 & \frac {1}{x} \end {array}\right ] \\ &= \frac {4}{x^{2}} \end {align*}
Now we are ready to evaluate each \(U_i(x)\). \begin {align*} U_1 &= (-1)^{4-1} \int { \frac {F(x) W_1(x) }{a W(x)} \, dx}\\ &= (-1)^{3} \int { \frac { \left (9 x^{2}\right ) \left (x\right )}{\left (x^{4}\right ) \left (\frac {12}{x^{3}}\right )} \, dx} \\ &= - \int { \frac {9 x^{3}}{12 x} \, dx}\\ &= - \int {\left (\frac {3 x^{2}}{4}\right ) \, dx}\\ &= -\frac {x^{3}}{4} \end {align*}
\begin {align*} U_2 &= (-1)^{4-2} \int { \frac {F(x) W_2(x) }{a W(x)} \, dx}\\ &= (-1)^{2} \int { \frac { \left (9 x^{2}\right ) \left (\frac {6 \ln \left (x \right )-3}{x}\right )}{\left (x^{4}\right ) \left (\frac {12}{x^{3}}\right )} \, dx} \\ &= \int { \frac {9 x \left (6 \ln \left (x \right )-3\right )}{12 x} \, dx}\\ &= \int {\left (\frac {9 \ln \left (x \right )}{2}-\frac {9}{4}\right ) \, dx}\\ &= -\frac {27 x}{4}+\frac {9 x \ln \left (x \right )}{2} \end {align*}
\begin {align*} U_3 &= (-1)^{4-3} \int { \frac {F(x) W_3(x) }{a W(x)} \, dx}\\ &= (-1)^{1} \int { \frac { \left (9 x^{2}\right ) \left (\frac {6}{x}\right )}{\left (x^{4}\right ) \left (\frac {12}{x^{3}}\right )} \, dx} \\ &= - \int { \frac {54 x}{12 x} \, dx}\\ &= - \int {\left ({\frac {9}{2}}\right ) \, dx}\\ &= -\frac {9 x}{2} \end {align*}
\begin {align*} U_4 &= (-1)^{4-4} \int { \frac {F(x) W_4(x) }{a W(x)} \, dx}\\ &= (-1)^{0} \int { \frac { \left (9 x^{2}\right ) \left (\frac {4}{x^{2}}\right )}{\left (x^{4}\right ) \left (\frac {12}{x^{3}}\right )} \, dx} \\ &= \int { \frac {36}{12 x} \, dx}\\ &= \int {\left (\frac {3}{x}\right ) \, dx}\\ &= 3 \ln \left (x \right ) \end {align*}
Now that all the \(U_i\) functions have been determined, the particular solution is found from \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3+U_4 y_4 \] Hence \begin {equation*} \begin {split} y_p &= \left (-\frac {x^{3}}{4}\right ) \left (\frac {1}{x}\right ) \\ &+\left (-\frac {27 x}{4}+\frac {9 x \ln \left (x \right )}{2}\right ) \left (x\right ) \\ &+\left (-\frac {9 x}{2}\right ) \left (x \ln \left (x \right )\right ) \\ &+\left (3 \ln \left (x \right )\right ) \left (x^{2}\right ) \end {split} \end {equation*} Therefore the particular solution is \[ y_p = x^{2} \left (-7+3 \ln \left (x \right )\right ) \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\frac {c_{1}}{x}+c_{2} x +\ln \left (x \right ) c_{3} x +c_{4} x^{2}\right ) + \left (x^{2} \left (-7+3 \ln \left (x \right )\right )\right ) \\ \end{align*} Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \frac {c_{1}}{x}+c_{2} x +\ln \left (x \right ) c_{3} x +c_{4} x^{2}+x^{2} \left (-7+3 \ln \left (x \right )\right ) \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 \(y = -7\) and \(x = 1\) in the above gives \begin {align*} -7 = c_{1} +c_{2} +c_{4} -7\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {c_{1}}{x^{2}}+c_{2} +c_{3} +\ln \left (x \right ) c_{3} +2 c_{4} x +2 x \left (-7+3 \ln \left (x \right )\right )+3 x \end {align*}
substituting \(y^{\prime } = -11\) and \(x = 1\) in the above gives \begin {align*} -11 = -c_{1} +c_{2} +c_{3} +2 c_{4} -11\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = \frac {2 c_{1}}{x^{3}}+\frac {c_{3}}{x}+2 c_{4} -5+6 \ln \left (x \right ) \end {align*}
substituting \(y^{\prime \prime } = -5\) and \(x = 1\) in the above gives \begin {align*} -5 = 2 c_{1} +c_{3} +2 c_{4} -5\tag {3A} \end {align*}
Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = -\frac {6 c_{1}}{x^{4}}-\frac {c_{3}}{x^{2}}+\frac {6}{x} \end {align*}
substituting \(y^{\prime \prime \prime } = 6\) and \(x = 1\) in the above gives \begin {align*} 6 = -6 c_{1} -c_{3} +6\tag {4A} \end {align*}
Equations {1A,2A,3A,4A} are now solved for \(\{c_{1}, c_{2}, c_{3}, c_{4}\}\). Solving for the constants gives \begin {align*} c_{1}&=0\\ c_{2}&=0\\ c_{3}&=0\\ c_{4}&=0 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = 3 \ln \left (x \right ) x^{2}-7 x^{2} \end {align*}
Which simplifies to \[ y = x^{2} \left (-7+3 \ln \left (x \right )\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= x^{2} \left (-7+3 \ln \left (x \right )\right ) \\ \end{align*}
Verification of solutions
\[ y = x^{2} \left (-7+3 \ln \left (x \right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{4} y^{\prime \prime \prime \prime }+3 y^{\prime \prime \prime } x^{3}-x^{2} y^{\prime \prime }+2 y^{\prime } x -2 y=9 x^{2}, y \left (1\right )=-7, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=-11, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=-5, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=6\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & y^{\prime \prime \prime \prime } \end {array} \]
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 -> Calling odsolve with the ODE`, diff(diff(diff(_b(_a), _a), _a), _a) = (c__1+2*_a*_b(_a)-2*(diff(_b(_a), _a))*_a^2+(diff(diff(_b(_ Methods for third order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable <- high order exact linear fully integrable successful <- high order exact_linear_nonhomogeneous successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 14
dsolve([x^4*diff(y(x),x$4)+3*x^3*diff(y(x),x$3)-x^2*diff(y(x),x$2)+2*x*diff(y(x),x)-2*y(x)=9*x^2,y(1) = -7, D(y)(1) = -11, (D@@2)(y)(1) = -5, (D@@3)(y)(1) = 6],y(x), singsol=all)
\[ y \left (x \right ) = x^{2} \left (-7+3 \ln \left (x \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.008 (sec). Leaf size: 15
DSolve[{x^4*y''''[x]+3*x^3*y'''[x]-x^2*y''[x]+2*x*y'[x]-2*y[x]==9*x^2,{y[1]==-7,y'[1]==-11,y''[1]==-5,y'''[1]==6}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x^2 (3 \log (x)-7) \]