Internal problem ID [1024]
Internal file name [OUTPUT/1025_Sunday_June_05_2022_01_57_12_AM_91127049/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 2, First order equations. Transformation of Nonlinear Equations into Separable
Equations. Section 2.4 Page 68
Problem number: 49.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(y)]`], _Riccati]
\[ \boxed {x \ln \left (x \right )^{2} y^{\prime }-\ln \left (x \right ) y-y^{2}=-4 \ln \left (x \right )^{2}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {-4 \ln \left (x \right )^{2}+\ln \left (x \right ) y +y^{2}}{x \ln \left (x \right )^{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\frac {4}{x}+\frac {y}{x \ln \left (x \right )}+\frac {y^{2}}{x \ln \left (x \right )^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {4}{x}\), \(f_1(x)=\frac {1}{x \ln \left (x \right )}\) and \(f_2(x)=\frac {1}{x \ln \left (x \right )^{2}}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{x \ln \left (x \right )^{2}}} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=-\frac {1}{x^{2} \ln \left (x \right )^{2}}-\frac {2}{x^{2} \ln \left (x \right )^{3}}\\ f_1 f_2 &=\frac {1}{x^{2} \ln \left (x \right )^{3}}\\ f_2^2 f_0 &=-\frac {4}{x^{3} \ln \left (x \right )^{4}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {u^{\prime \prime }\left (x \right )}{x \ln \left (x \right )^{2}}-\left (-\frac {1}{x^{2} \ln \left (x \right )^{2}}-\frac {1}{x^{2} \ln \left (x \right )^{3}}\right ) u^{\prime }\left (x \right )-\frac {4 u \left (x \right )}{x^{3} \ln \left (x \right )^{4}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \frac {c_{2} \ln \left (x \right )^{4}+c_{1}}{\ln \left (x \right )^{2}} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {2 c_{2} \ln \left (x \right )^{4}-2 c_{1}}{x \ln \left (x \right )^{3}} \] Using the above in (1) gives the solution \[ y = -\frac {\left (2 c_{2} \ln \left (x \right )^{4}-2 c_{1} \right ) \ln \left (x \right )}{c_{2} \ln \left (x \right )^{4}+c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {2 \left (c_{2} \ln \left (x \right )^{4}-c_{1} \right ) \ln \left (x \right )}{c_{2} \ln \left (x \right )^{4}+c_{1}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {2 \left (c_{2} \ln \left (x \right )^{4}-c_{1} \right ) \ln \left (x \right )}{c_{2} \ln \left (x \right )^{4}+c_{1}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {2 \left (c_{2} \ln \left (x \right )^{4}-c_{1} \right ) \ln \left (x \right )}{c_{2} \ln \left (x \right )^{4}+c_{1}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \ln \left (x \right )^{2} y^{\prime }-\ln \left (x \right ) y-y^{2}=-4 \ln \left (x \right )^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-4 \ln \left (x \right )^{2}+\ln \left (x \right ) y+y^{2}}{x \ln \left (x \right )^{2}} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini <- Chini successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 18
dsolve(x*(ln(x))^2*diff(y(x),x)=-4*(ln(x))^2+y(x)*ln(x)+y(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = 2 i \tan \left (2 i \ln \left (\ln \left (x \right )\right )+c_{1} \right ) \ln \left (x \right ) \]
✓ Solution by Mathematica
Time used: 1.259 (sec). Leaf size: 64
DSolve[x*(Log[x])^2*y'[x]==-4*(Log[x])^2+y[x]*Log[x]+y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to 2 i \log (x) \tan (2 i \log (\log (x))+c_1) \\ y(x)\to \frac {2 \log (x) \left (-\log ^4(x)+e^{2 i \text {Interval}[\{0,\pi \}]}\right )}{\log ^4(x)+e^{2 i \text {Interval}[\{0,\pi \}]}} \\ \end{align*}