Internal problem ID [14705]
Internal file name [OUTPUT/14385_Wednesday_April_03_2024_02_17_56_PM_85428939/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.6, page 187
Problem number: 31.
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]]
\[ \boxed {t^{2} \ln \left (t \right ) y^{\prime \prime \prime }-t y^{\prime \prime }+y^{\prime }=1} \] Since \(y\) is missing from the ode then we can use the substitution \(y^{\prime } = v \left (t \right )\) to reduce the order by one. The ODE becomes \begin {align*} t^{2} \ln \left (t \right ) v^{\prime \prime }\left (t \right )-v^{\prime }\left (t \right ) t +v \left (t \right ) = 0 \end {align*}
In normal form the ode \begin {align*} t^{2} \ln \left (t \right ) v^{\prime \prime }\left (t \right )-v^{\prime }\left (t \right ) t +v \left (t \right )&=0 \tag {1} \end {align*}
Becomes \begin {align*} v^{\prime \prime }\left (t \right )+p \left (t \right ) v^{\prime }\left (t \right )+q \left (t \right ) v \left (t \right )&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (t \right )&=-\frac {1}{t \ln \left (t \right )}\\ q \left (t \right )&=\frac {1}{t^{2} \ln \left (t \right )} \end {align*}
Applying change of variables on the depndent variable \(v \left (t \right ) = v \left (t \right ) t^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (t \right )\) and not \(v \left (t \right )\). \begin {align*} v^{\prime \prime }\left (t \right )+\left (\frac {2 n}{t}+p \right ) v^{\prime }\left (t \right )+\left (\frac {n \left (n -1\right )}{t^{2}}+\frac {n p}{t}+q \right ) v \left (t \right )&=0 \tag {3} \end {align*}
Let the coefficient of \(v \left (t \right )\) above be zero. Hence \begin {align*} \frac {n \left (n -1\right )}{t^{2}}+\frac {n p}{t}+q&=0 \tag {4} \end {align*}
Substituting the earlier values found for \(p \left (t \right )\) and \(q \left (t \right )\) into (4) gives \begin {align*} \frac {n \left (n -1\right )}{t^{2}}-\frac {n}{t^{2} \ln \left (t \right )}+\frac {1}{t^{2} \ln \left (t \right )}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=1 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (t \right )+\left (\frac {2}{t}-\frac {1}{t \ln \left (t \right )}\right ) v^{\prime }\left (t \right )&=0 \\ v^{\prime \prime }\left (t \right )+\left (\frac {2}{t}-\frac {1}{t \ln \left (t \right )}\right ) v^{\prime }\left (t \right )&=0 \tag {7} \\ \end {align*}
Using the substitution \begin {align*} u \left (t \right ) = v^{\prime }\left (t \right ) \end {align*}
Then (7) becomes \begin {align*} u^{\prime }\left (t \right )+\left (\frac {2}{t}-\frac {1}{t \ln \left (t \right )}\right ) u \left (t \right ) = 0 \tag {8} \\ \end {align*}
The above is now solved for \(u \left (t \right )\). In canonical form the ODE is \begin {align*} u' &= F(t,u)\\ &= f( t) g(u)\\ &= -\frac {u \left (-1+2 \ln \left (t \right )\right )}{t \ln \left (t \right )} \end {align*}
Where \(f(t)=-\frac {-1+2 \ln \left (t \right )}{t \ln \left (t \right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {-1+2 \ln \left (t \right )}{t \ln \left (t \right )} \,d t\\ \int { \frac {1}{u} \,du} &= \int {-\frac {-1+2 \ln \left (t \right )}{t \ln \left (t \right )} \,d t}\\ \ln \left (u \right )&=-2 \ln \left (t \right )+\ln \left (\ln \left (t \right )\right )+c_{1}\\ u&={\mathrm e}^{-2 \ln \left (t \right )+\ln \left (\ln \left (t \right )\right )+c_{1}}\\ &=c_{1} {\mathrm e}^{-2 \ln \left (t \right )+\ln \left (\ln \left (t \right )\right )} \end {align*}
Which simplifies to \[ u \left (t \right ) = \frac {c_{1} \ln \left (t \right )}{t^{2}} \] Now that \(u \left (t \right )\) is known, then \begin {align*} v^{\prime }\left (t \right )&= u \left (t \right )\\ v \left (t \right )&= \int u \left (t \right )d t +c_{2}\\ &= c_{1} \left (-\frac {\ln \left (t \right )}{t}-\frac {1}{t}\right )+c_{2} \end {align*}
Hence \begin {align*} v \left (t \right )&= v \left (t \right ) t^{n}\\ &= \left (c_{1} \left (-\frac {\ln \left (t \right )}{t}-\frac {1}{t}\right )+c_{2} \right ) t\\ &= -c_{1} \ln \left (t \right )+c_{2} t -c_{1}\\ \end {align*}
But since \(y^{\prime } = v \left (t \right )\) then we now need to solve the ode \(y^{\prime } = \left (c_{1} \left (-\frac {\ln \left (t \right )}{t}-\frac {1}{t}\right )+c_{2} \right ) t\). Integrating both sides gives \begin {align*} y &= \int { -c_{1} \ln \left (t \right )+c_{2} t -c_{1}\,\mathop {\mathrm {d}t}}\\ &= \frac {c_{2} t^{2}}{2}-\ln \left (t \right ) c_{1} t +c_{3} \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 \[ t^{2} \ln \left (t \right ) y^{\prime \prime \prime }-t y^{\prime \prime }+y^{\prime } = 0 \] 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 \(U_i(t)\) are functions to be determined as follows \[ U_i = (-1)^{n-i} \int { \frac {F(t) W_i(t) }{a W(t)} \, dt} \] Where \(W(t)\) is the Wronskian and \(W_i(t)\) 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 \(F(t)\) is the RHS of the ODE. Therefore, the first step is to find the Wronskian \(W \left (t \right )\). This is given by \begin {equation*} W(t) = \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 & t^{2} & \ln \left (t \right ) t \\ 0 & 2 t & \ln \left (t \right )+1 \\ 0 & 2 & \frac {1}{t} \end {array}\right ] \\ |W| &= -2 \ln \left (t \right ) \end {align*}
The determinant simplifies to \begin {align*} |W| &= -2 \ln \left (t \right ) \end {align*}
Now we determine \(W_i\) for each \(U_i\). \begin {align*} W_1(t) &= \det \,\left [\begin {array}{cc} t^{2} & \ln \left (t \right ) t \\ 2 t & \ln \left (t \right )+1 \end {array}\right ] \\ &= -t^{2} \left (-1+\ln \left (t \right )\right ) \end {align*}
\begin {align*} W_2(t) &= \det \,\left [\begin {array}{cc} 1 & \ln \left (t \right ) t \\ 0 & \ln \left (t \right )+1 \end {array}\right ] \\ &= \ln \left (t \right )+1 \end {align*}
\begin {align*} W_3(t) &= \det \,\left [\begin {array}{cc} 1 & t^{2} \\ 0 & 2 t \end {array}\right ] \\ &= 2 t \end {align*}
Now we are ready to evaluate each \(U_i(t)\). \begin {align*} U_1 &= (-1)^{3-1} \int { \frac {F(t) W_1(t) }{a W(t)} \, dt}\\ &= (-1)^{2} \int { \frac { \left (1\right ) \left (-t^{2} \left (-1+\ln \left (t \right )\right )\right )}{\left (t^{2} \ln \left (t \right )\right ) \left (-2 \ln \left (t \right )\right )} \, dt} \\ &= \int { \frac {-t^{2} \left (-1+\ln \left (t \right )\right )}{-2 t^{2} \ln \left (t \right )^{2}} \, dt}\\ &= \int {\left (\frac {-1+\ln \left (t \right )}{2 \ln \left (t \right )^{2}}\right ) \, dt}\\ &= \frac {t}{2 \ln \left (t \right )} \end {align*}
\begin {align*} U_2 &= (-1)^{3-2} \int { \frac {F(t) W_2(t) }{a W(t)} \, dt}\\ &= (-1)^{1} \int { \frac { \left (1\right ) \left (\ln \left (t \right )+1\right )}{\left (t^{2} \ln \left (t \right )\right ) \left (-2 \ln \left (t \right )\right )} \, dt} \\ &= - \int { \frac {\ln \left (t \right )+1}{-2 t^{2} \ln \left (t \right )^{2}} \, dt}\\ &= - \int {\left (-\frac {\ln \left (t \right )+1}{2 t^{2} \ln \left (t \right )^{2}}\right ) \, dt}\\ &= -\frac {1}{2 t \ln \left (t \right )} \end {align*}
\begin {align*} U_3 &= (-1)^{3-3} \int { \frac {F(t) W_3(t) }{a W(t)} \, dt}\\ &= (-1)^{0} \int { \frac { \left (1\right ) \left (2 t\right )}{\left (t^{2} \ln \left (t \right )\right ) \left (-2 \ln \left (t \right )\right )} \, dt} \\ &= \int { \frac {2 t}{-2 t^{2} \ln \left (t \right )^{2}} \, dt}\\ &= \int {\left (-\frac {1}{\ln \left (t \right )^{2} t}\right ) \, dt}\\ &= \frac {1}{\ln \left (t \right )} \end {align*}
Now that all the \(U_i\) functions have been determined, the particular solution is found from \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3 \] Hence \begin {equation*} \begin {split} y_p &= \left (\frac {t}{2 \ln \left (t \right )}\right ) \\ &+\left (-\frac {1}{2 t \ln \left (t \right )}\right ) \left (t^{2}\right ) \\ &+\left (\frac {1}{\ln \left (t \right )}\right ) \left (\ln \left (t \right ) t\right ) \end {split} \end {equation*} Therefore the particular solution is \[ y_p = t \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (y &= \frac {c_{2} t^{2}}{2}-\ln \left (t \right ) c_{1} t +c_{3}\right ) + \left (t\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{2} t^{2}}{2}-\ln \left (t \right ) c_{1} t +c_{3} +t \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{2} t^{2}}{2}-\ln \left (t \right ) c_{1} t +c_{3} +t \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t^{2} \ln \left (t \right ) y^{\prime \prime \prime }-y^{\prime \prime } t +y^{\prime }=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \end {array} \]
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] -> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = -(-(diff(_b(_a), _a))*_a+_b(_a)-1)/(ln(_a)*_a^2), _b(_a)` *** Suble 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)] trying symmetries linear in x and y(x) <- linear symmetries successful <- differential order: 3; linear nonhomogeneous with symmetry [0,1] successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 19
dsolve(t^2*ln(t)*diff(y(t),t$3)-t*diff(y(t),t$2)+diff(y(t),t)=1,y(t), singsol=all)
\[ y \left (t \right ) = \frac {c_{2} t^{2}}{2}-\ln \left (t \right ) c_{1} t +t +c_{3} \]
✓ Solution by Mathematica
Time used: 0.082 (sec). Leaf size: 25
DSolve[t^2*Log[t]*y'''[t]-t*y''[t]+y'[t]==1,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {c_1 t^2}{2}+t-c_2 t \log (t)+c_3 \]