Internal problem ID [11286]
Internal file name [OUTPUT/10272_Wednesday_December_21_2022_03_47_18_PM_2332710/index.tex]
Book: An elementary treatise on differential equations by Abraham Cohen. DC heath publishers.
1906
Section: Chapter VII, Linear differential equations with constant coefficients. Article 52.
Summary. Page 117
Problem number: Ex 14.
ODE order: 3.
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
[[_3rd_order, _exact, _linear, _nonhomogeneous]]
\[ \boxed {x^{3} y^{\prime \prime \prime }+2 x^{2} y^{\prime \prime }-y^{\prime } x +y=\frac {1}{x}} \] 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 \[ x^{3} y^{\prime \prime \prime }+2 x^{2} y^{\prime \prime }-y^{\prime } x +y = 0 \] 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} \end {align*}
Substituting these back into \[ x^{3} y^{\prime \prime \prime }+2 x^{2} y^{\prime \prime }-y^{\prime } x +y = \frac {1}{x} \] gives \[ -x \lambda \,x^{\lambda -1}+2 x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}+x^{\lambda } = 0 \] Which simplifies to \[ -\lambda \,x^{\lambda }+2 \lambda \left (\lambda -1\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }+x^{\lambda } = 0 \] And since \(x^{\lambda }\neq 0\) then dividing through by \(x^{\lambda }\), the above becomes
\[ -\lambda +2 \lambda \left (\lambda -1\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right )+1 = 0 \] Simplifying gives the characteristic equation as \[ \left (\lambda +1\right ) \left (\lambda -1\right )^{2} = 0 \] Solving the above gives the following roots \begin {align*} \lambda _1 &= -1\\ \lambda _2 &= 1\\ \lambda _3 &= 1 \end {align*}
This table summarises the result
| root | multiplicity | type of root |
| \(-1\) | \(1\) | real root |
| \(1\) | \(2\) | 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
\[ y = \frac {c_{1}}{x}+c_{2} x +c_{3} \ln \left (x \right ) x \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= \frac {1}{x} \\ y_2 &= x \\ y_3 &= \ln \left (x \right ) x \\ \end{align*} Now the particular solution to the given ODE is found \[ x^{3} y^{\prime \prime \prime }+2 x^{2} y^{\prime \prime }-y^{\prime } x +y = \frac {1}{x} \] Let the particular solution be \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3 \] 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_1'&y_2'&y_3'\\ y_1''&y_2''&y_3''\\ \end {vmatrix} \end {equation*} Substituting the fundamental set of solutions \(y_i\) found above in the Wronskian gives \begin {align*} W &= \left [\begin {array}{ccc} \frac {1}{x} & x & \ln \left (x \right ) x \\ -\frac {1}{x^{2}} & 1 & 1+\ln \left (x \right ) \\ \frac {2}{x^{3}} & 0 & \frac {1}{x} \end {array}\right ] \\ |W| &= \frac {4}{x^{2}} \end {align*}
The determinant simplifies to \begin {align*} |W| &= \frac {4}{x^{2}} \end {align*}
Now we determine \(W_i\) for each \(U_i\). \begin {align*} W_1(x) &= \det \,\left [\begin {array}{cc} x & \ln \left (x \right ) x \\ 1 & 1+\ln \left (x \right ) \end {array}\right ] \\ &= x \end {align*}
\begin {align*} W_2(x) &= \det \,\left [\begin {array}{cc} \frac {1}{x} & \ln \left (x \right ) x \\ -\frac {1}{x^{2}} & 1+\ln \left (x \right ) \end {array}\right ] \\ &= \frac {1+2 \ln \left (x \right )}{x} \end {align*}
\begin {align*} W_3(x) &= \det \,\left [\begin {array}{cc} \frac {1}{x} & x \\ -\frac {1}{x^{2}} & 1 \end {array}\right ] \\ &= \frac {2}{x} \end {align*}
Now we are ready to evaluate each \(U_i(x)\). \begin {align*} U_1 &= (-1)^{3-1} \int { \frac {F(x) W_1(x) }{a W(x)} \, dx}\\ &= (-1)^{2} \int { \frac { \left (\frac {1}{x}\right ) \left (x\right )}{\left (x^{3}\right ) \left (\frac {4}{x^{2}}\right )} \, dx} \\ &= \int { \frac {1}{4 x} \, dx}\\ &= \int {\left (\frac {1}{4 x}\right ) \, dx}\\ &= \frac {\ln \left (x \right )}{4} \end {align*}
\begin {align*} U_2 &= (-1)^{3-2} \int { \frac {F(x) W_2(x) }{a W(x)} \, dx}\\ &= (-1)^{1} \int { \frac { \left (\frac {1}{x}\right ) \left (\frac {1+2 \ln \left (x \right )}{x}\right )}{\left (x^{3}\right ) \left (\frac {4}{x^{2}}\right )} \, dx} \\ &= - \int { \frac {\frac {1+2 \ln \left (x \right )}{x^{2}}}{4 x} \, dx}\\ &= - \int {\left (\frac {1+2 \ln \left (x \right )}{4 x^{3}}\right ) \, dx}\\ &= \frac {1}{4 x^{2}}+\frac {\ln \left (x \right )}{4 x^{2}} \end {align*}
\begin {align*} U_3 &= (-1)^{3-3} \int { \frac {F(x) W_3(x) }{a W(x)} \, dx}\\ &= (-1)^{0} \int { \frac { \left (\frac {1}{x}\right ) \left (\frac {2}{x}\right )}{\left (x^{3}\right ) \left (\frac {4}{x^{2}}\right )} \, dx} \\ &= \int { \frac {\frac {2}{x^{2}}}{4 x} \, dx}\\ &= \int {\left (\frac {1}{2 x^{3}}\right ) \, dx}\\ &= -\frac {1}{4 x^{2}} \end {align*}
Now that all the \(U_i\) functions have been determined, the particular solution is found from \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3 \] Hence \begin {equation*} \begin {split} y_p &= \left (\frac {\ln \left (x \right )}{4}\right ) \left (\frac {1}{x}\right ) \\ &+\left (\frac {1}{4 x^{2}}+\frac {\ln \left (x \right )}{4 x^{2}}\right ) \left (x\right ) \\ &+\left (-\frac {1}{4 x^{2}}\right ) \left (\ln \left (x \right ) x\right ) \end {split} \end {equation*} Therefore the particular solution is \[ y_p = \frac {1+\ln \left (x \right )}{4 x} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\frac {c_{1}}{x}+c_{2} x +c_{3} \ln \left (x \right ) x\right ) + \left (\frac {1+\ln \left (x \right )}{4 x}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1}}{x}+c_{2} x +c_{3} \ln \left (x \right ) x +\frac {1+\ln \left (x \right )}{4 x} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1}}{x}+c_{2} x +c_{3} \ln \left (x \right ) x +\frac {1+\ln \left (x \right )}{4 x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )+2 x^{2} \left (\frac {d}{d x}y^{\prime }\right )-y^{\prime } x +y=\frac {1}{x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d}{d x}y^{\prime \prime } \end {array} \]
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 3; linear nonhomogeneous with symmetry [0,1] trying high order linear exact nonhomogeneous -> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = (c__1-_b(_a)*_a+_a^2*(diff(_b(_a), _a))+ln(_a))/_a^3, _b(_a)` *** S Methods for second 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: 0.0 (sec). Leaf size: 28
dsolve(x^3*diff(y(x),x$3)+2*x^2*diff(y(x),x$2)-x*diff(y(x),x)+y(x)=1/x,y(x), singsol=all)
\[ y \left (x \right ) = \frac {4 \ln \left (x \right ) c_{2} x^{2}+4 c_{3} x^{2}+\ln \left (x \right )+c_{1} +1}{4 x} \]
✓ Solution by Mathematica
Time used: 0.013 (sec). Leaf size: 33
DSolve[x^3*y'''[x]+2*x^2*y''[x]-x*y'[x]+y[x]==1/x,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {\log (x)+1}{4 x}+\frac {c_1}{x}+c_2 x+c_3 x \log (x) \]