Internal problem ID [13488]
Internal file name [OUTPUT/12660_Friday_February_16_2024_12_04_46_AM_50008945/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 13. Higher order equations: Extending first order concepts. Additional exercises
page 259
Problem number: 13.3 (b).
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_missing_y"
Maple gives the following as the ode type
[[_3rd_order, _missing_y]]
\[ \boxed {x y^{\prime \prime \prime }+2 y^{\prime \prime }=6 x} \] Since \(y\) is missing from the ode then we can use the substitution \(y^{\prime } = v \left (x \right )\) to reduce the order by one. The ODE becomes \begin {align*} x v^{\prime \prime }\left (x \right )+2 v^{\prime }\left (x \right ) = 0 \end {align*}
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (x v^{\prime \prime }\left (x \right )+2 v^{\prime }\left (x \right )\right )d x &= 0 \\ v^{\prime }\left (x \right ) x +v \left (x \right ) = c_{1} \end {align*}
Which is now solved for \(v \left (x \right )\). In canonical form the ODE is \begin {align*} v' &= F(x,v)\\ &= f( x) g(v)\\ &= \frac {-v +c_{1}}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(v)=-v +c_{1}\). Integrating both sides gives \begin{align*} \frac {1}{-v +c_{1}} \,dv &= \frac {1}{x} \,d x \\ \int { \frac {1}{-v +c_{1}} \,dv} &= \int {\frac {1}{x} \,d x} \\ -\ln \left (v -c_{1} \right )&=\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \frac {1}{v -c_{1}} &= {\mathrm e}^{\ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \frac {1}{v -c_{1}} &= c_{3} x \end {align*}
Which simplifies to \[ v \left (x \right ) = \frac {\left (c_{3} {\mathrm e}^{c_{2}} x c_{1} +1\right ) {\mathrm e}^{-c_{2}}}{c_{3} x} \] But since \(y^{\prime } = v \left (x \right )\) then we now need to solve the ode \(y^{\prime } = \frac {\left (c_{3} {\mathrm e}^{c_{2}} x c_{1} +1\right ) {\mathrm e}^{-c_{2}}}{c_{3} x}\). Integrating both sides gives \begin {align*} y &= \int { \frac {\left (c_{3} {\mathrm e}^{c_{2}} x c_{1} +1\right ) {\mathrm e}^{-c_{2}}}{c_{3} x}\,\mathop {\mathrm {d}x}}\\ &= c_{1} x +\frac {{\mathrm e}^{-c_{2}} \ln \left (x \right )}{c_{3}}+c_{4} \end {align*}
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 \[ x y^{\prime \prime \prime }+2 y^{\prime \prime } = 0 \] 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} 1 & x & \ln \left (x \right ) \\ 0 & 1 & \frac {1}{x} \\ 0 & 0 & -\frac {1}{x^{2}} \end {array}\right ] \\ |W| &= -\frac {1}{x^{2}} \end {align*}
The determinant simplifies to \begin {align*} |W| &= -\frac {1}{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 ) \\ 1 & \frac {1}{x} \end {array}\right ] \\ &= 1-\ln \left (x \right ) \end {align*}
\begin {align*} W_2(x) &= \det \,\left [\begin {array}{cc} 1 & \ln \left (x \right ) \\ 0 & \frac {1}{x} \end {array}\right ] \\ &= \frac {1}{x} \end {align*}
\begin {align*} W_3(x) &= \det \,\left [\begin {array}{cc} 1 & x \\ 0 & 1 \end {array}\right ] \\ &= 1 \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 (6 x\right ) \left (1-\ln \left (x \right )\right )}{\left (x\right ) \left (-\frac {1}{x^{2}}\right )} \, dx} \\ &= \int { \frac {6 x \left (1-\ln \left (x \right )\right )}{-\frac {1}{x}} \, dx}\\ &= \int {\left (-6 x^{2} \left (1-\ln \left (x \right )\right )\right ) \, dx}\\ &= -\frac {8 x^{3}}{3}+2 x^{3} \ln \left (x \right ) \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 (6 x\right ) \left (\frac {1}{x}\right )}{\left (x\right ) \left (-\frac {1}{x^{2}}\right )} \, dx} \\ &= - \int { \frac {6}{-\frac {1}{x}} \, dx}\\ &= - \int {\left (-6 x\right ) \, dx}\\ &= 3 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 (6 x\right ) \left (1\right )}{\left (x\right ) \left (-\frac {1}{x^{2}}\right )} \, dx} \\ &= \int { \frac {6 x}{-\frac {1}{x}} \, dx}\\ &= \int {\left (-6 x^{2}\right ) \, dx}\\ &= -2 x^{3} \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 {8 x^{3}}{3}+2 x^{3} \ln \left (x \right )\right ) \\ &+\left (3 x^{2}\right ) \left (x\right ) \\ &+\left (-2 x^{3}\right ) \left (\ln \left (x \right )\right ) \end {split} \end {equation*} Therefore the particular solution is \[ y_p = \frac {x^{3}}{3} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (y &= c_{1} x +\frac {{\mathrm e}^{-c_{2}} \ln \left (x \right )}{c_{3}}+c_{4}\right ) + \left (\frac {x^{3}}{3}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x +\frac {{\mathrm e}^{-c_{2}} \ln \left (x \right )}{c_{3}}+c_{4} +\frac {x^{3}}{3} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x +\frac {{\mathrm e}^{-c_{2}} \ln \left (x \right )}{c_{3}}+c_{4} +\frac {x^{3}}{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime \prime }\right )+2 \frac {d}{d x}y^{\prime }=6 x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d}{d x}y^{\prime \prime } \end {array} \]
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable -> Calling odsolve with the ODE`, diff(_b(_a), _a) = -2*(_b(_a)-3*_a)/_a, _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful <- high order exact linear fully integrable successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 19
dsolve(x*diff(y(x),x$3)+2*diff(y(x),x$2)=6*x,y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{3}}{3}-c_{1} \ln \left (x \right )+c_{2} x +c_{3} \]
✓ Solution by Mathematica
Time used: 0.048 (sec). Leaf size: 25
DSolve[x*y'''[x]+2*y''[x]==6*x,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {x^3}{3}+c_3 x-c_1 \log (x)+c_2 \]