Internal problem ID [9110]
Internal file name [OUTPUT/8045_Monday_June_06_2022_01_22_20_AM_12992046/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 776.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati", "bernoulli", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_Bernoulli]
\[ \boxed {y^{\prime }-\frac {y \left (-\ln \left (\frac {1}{x}\right )-\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {x^{2}+1}{x}\right ) x^{2} y\right )}{x \ln \left (\frac {1}{x}\right )}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {y \left (-\ln \left (\frac {1}{x}\right )-\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {x^{2}+1}{x}\right ) x^{2} y \right )}{x \ln \left (\frac {1}{x}\right )}\\ y^{\prime }&= \omega \left ( x,y\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end {align*}
The type of this ode is known. It is of type Bernoulli. Therefore we do not need
to solve the PDE (A), and can just use the lookup table shown below to find \(\xi ,\eta \)
|
ODE class |
Form |
\(\xi \) |
\(\eta \) |
|
linear ode |
\(y'=f(x) y(x) +g(x)\) |
\(0\) |
\(e^{\int fdx}\) |
|
separable ode |
\(y^{\prime }=f\left ( x\right ) g\left ( y\right ) \) |
\(\frac {1}{f}\) |
\(0\) |
|
quadrature ode |
\(y^{\prime }=f\left ( x\right ) \) |
\(0\) |
\(1\) |
|
quadrature ode |
\(y^{\prime }=g\left ( y\right ) \) |
\(1\) |
\(0\) |
|
homogeneous ODEs of Class A |
\(y^{\prime }=f\left ( \frac {y}{x}\right ) \) |
\(x\) |
\(y\) |
|
homogeneous ODEs of Class C |
\(y^{\prime }=\left ( a+bx+cy\right ) ^{\frac {n}{m}}\) |
\(1\) |
\(-\frac {b}{c}\) |
|
homogeneous class D |
\(y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left (\frac {y}{x}\right ) \) |
\(x^{2}\) |
\(xy\) |
|
First order special form ID 1 |
\(y^{\prime }=g\left ( x\right ) e^{h\left (x\right ) +by}+f\left ( x\right ) \) |
\(\frac {e^{-\int bf\left ( x\right )dx-h\left ( x\right ) }}{g\left ( x\right ) }\) |
\(\frac {f\left ( x\right )e^{-\int bf\left ( x\right ) dx-h\left ( x\right ) }}{g\left ( x\right ) }\) |
|
polynomial type ode |
\(y^{\prime }=\frac {a_{1}x+b_{1}y+c_{1}}{a_{2}x+b_{2}y+c_{2}}\) |
\(\frac {a_{1}b_{2}x-a_{2}b_{1}x-b_{1}c_{2}+b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) |
\(\frac {a_{1}b_{2}y-a_{2}b_{1}y-a_{1}c_{2}-a_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) |
|
Bernoulli ode |
\(y^{\prime }=f\left ( x\right ) y+g\left ( x\right ) y^{n}\) |
\(0\) |
\(e^{-\int \left ( n-1\right ) f\left ( x\right ) dx}y^{n}\) |
|
Reduced Riccati |
\(y^{\prime }=f_{1}\left ( x\right ) y+f_{2}\left ( x\right )y^{2}\) |
\(0\) |
\(e^{-\int f_{1}dx}\) |
|
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|
|
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The above table shows that \begin {align*} \xi \left (x,y\right ) &=0\\ \tag {A1} \eta \left (x,y\right ) &=y^{2} {\mathrm e}^{\int -\frac {-\ln \left (\frac {x^{2}+1}{x}\right ) x -\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x} \end {align*}
The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since \(\xi =0\) then in this special case \begin {align*} R = x \end {align*}
\(S\) is found from \begin {align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{y^{2} {\mathrm e}^{\int -\frac {-\ln \left (\frac {x^{2}+1}{x}\right ) x -\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x}}} dy \end {align*}
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y \left (-\ln \left (\frac {1}{x}\right )-\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {x^{2}+1}{x}\right ) x^{2} y \right )}{x \ln \left (\frac {1}{x}\right )} \end {align*}
This is a Bernoulli ODE. \[ y' = \frac {-\ln \left (\frac {x^{2}+1}{x}\right ) x -\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )} y +\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x}{\ln \left (\frac {1}{x}\right )} y^{2} \tag {1} \] The standard Bernoulli ODE has the form \[ y' = f_0(x)y+f_1(x)y^n \tag {2} \] The first step is to divide the above equation by \(y^n \) which gives \[ \frac {y'}{y^n} = f_0(x) y^{1-n} +f_1(x) \tag {3} \] The next step is use the substitution \(w = y^{1-n}\) in equation (3) which generates a new ODE in \(w \left (x \right )\) which will be linear and can be easily solved using an integrating factor. Backsubstitution then gives the solution \(y(x)\) which is what we want.
This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that \begin {align*} f_0(x)&=\frac {-\ln \left (\frac {x^{2}+1}{x}\right ) x -\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}\\ f_1(x)&=\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x}{\ln \left (\frac {1}{x}\right )}\\ n &=2 \end {align*}
Dividing both sides of ODE (1) by \(y^n=y^{2}\) gives \begin {align*} y'\frac {1}{y^{2}} &= \frac {-\ln \left (\frac {x^{2}+1}{x}\right ) x -\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right ) y} +\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x}{\ln \left (\frac {1}{x}\right )} \tag {4} \end {align*}
Let \begin {align*} w &= y^{1-n} \\ &= \frac {1}{y} \tag {5} \end {align*}
Taking derivative of equation (5) w.r.t \(x\) gives \begin {align*} w' &= -\frac {1}{y^{2}}y' \tag {6} \end {align*}
Substituting equations (5) and (6) into equation (4) gives \begin {align*} -w^{\prime }\left (x \right )&= \frac {\left (-\ln \left (\frac {x^{2}+1}{x}\right ) x -\ln \left (\frac {1}{x}\right )\right ) w \left (x \right )}{x \ln \left (\frac {1}{x}\right )}+\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x}{\ln \left (\frac {1}{x}\right )}\\ w' &= -\frac {\left (-\ln \left (\frac {x^{2}+1}{x}\right ) x -\ln \left (\frac {1}{x}\right )\right ) w}{x \ln \left (\frac {1}{x}\right )}-\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x}{\ln \left (\frac {1}{x}\right )} \tag {7} \end {align*}
The above now is a linear ODE in \(w \left (x \right )\) which is now solved.
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} w^{\prime }\left (x \right ) + p(x)w \left (x \right ) &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}\\ q(x) &=-\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x}{\ln \left (\frac {1}{x}\right )} \end {align*}
Hence the ode is \begin {align*} w^{\prime }\left (x \right )-\frac {\left (\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )\right ) w \left (x \right )}{x \ln \left (\frac {1}{x}\right )} = -\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x}{\ln \left (\frac {1}{x}\right )} \end {align*}
The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int -\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu w\right ) &= \left (\mu \right ) \left (-\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x}{\ln \left (\frac {1}{x}\right )}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left ({\mathrm e}^{\int -\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x} w\right ) &= \left ({\mathrm e}^{\int -\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x}\right ) \left (-\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x}{\ln \left (\frac {1}{x}\right )}\right )\\ \mathrm {d} \left ({\mathrm e}^{\int -\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x} w\right ) &= \left (-\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x \,{\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x \right )}}{\ln \left (\frac {1}{x}\right )}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{\int -\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x} w &= \int {-\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x \,{\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x \right )}}{\ln \left (\frac {1}{x}\right )}\,\mathrm {d} x}\\ {\mathrm e}^{\int -\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x} w &= \int -\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x \,{\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x \right )}}{\ln \left (\frac {1}{x}\right )}d x + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int -\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x}\) results in \begin {align*} w \left (x \right ) &= {\mathrm e}^{\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x} \left (\int -\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x \,{\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x \right )}}{\ln \left (\frac {1}{x}\right )}d x \right )+c_{1} {\mathrm e}^{\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x} \end {align*}
which simplifies to \begin {align*} w \left (x \right ) &= {\mathrm e}^{\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x} \left (-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x \,{\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x \right )}}{\ln \left (\frac {1}{x}\right )}d x \right )+c_{1} \right ) \end {align*}
Which can be simplified to become \[ w \left (x \right ) = {\mathrm e}^{\int \left (\frac {\ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}+\frac {1}{x}\right )d x} \left (-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x \,{\mathrm e}^{\int \left (-\frac {\ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}-\frac {1}{x}\right )d x}}{\ln \left (\frac {1}{x}\right )}d x \right )+c_{1} \right ) \]
Replacing \(w\) in the above by \(\frac {1}{y}\) using equation (5) gives the final solution. \begin {align*} \frac {1}{y} = {\mathrm e}^{\int \left (\frac {\ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}+\frac {1}{x}\right )d x} \left (-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x \,{\mathrm e}^{\int \left (-\frac {\ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}-\frac {1}{x}\right )d x}}{\ln \left (\frac {1}{x}\right )}d x \right )+c_{1} \right ) \end {align*}
Or \begin {align*} y = \frac {{\mathrm e}^{\int \left (-\frac {\ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}-\frac {1}{x}\right )d x}}{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x \,{\mathrm e}^{\int \left (-\frac {\ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}-\frac {1}{x}\right )d x}}{\ln \left (\frac {1}{x}\right )}d x \right )+c_{1}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{\int \left (-\frac {\ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}-\frac {1}{x}\right )d x}}{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x \,{\mathrm e}^{\int \left (-\frac {\ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}-\frac {1}{x}\right )d x}}{\ln \left (\frac {1}{x}\right )}d x \right )+c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \frac {{\mathrm e}^{\int \left (-\frac {\ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}-\frac {1}{x}\right )d x}}{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x \,{\mathrm e}^{\int \left (-\frac {\ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}-\frac {1}{x}\right )d x}}{\ln \left (\frac {1}{x}\right )}d x \right )+c_{1}} \] Verified OK.
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y \left (-\ln \left (\frac {1}{x}\right )-\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {x^{2}+1}{x}\right ) x^{2} y \right )}{x \ln \left (\frac {1}{x}\right )} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {y^{2} x \ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}-\frac {y \ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}-\frac {y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=0\), \(f_1(x)=\frac {-\ln \left (\frac {x^{2}+1}{x}\right ) x -\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}\) and \(f_2(x)=\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x}{\ln \left (\frac {1}{x}\right )}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {\ln \left (\frac {x^{2}+1}{x}\right ) x u}{\ln \left (\frac {1}{x}\right )}} \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 {\left (2-\frac {x^{2}+1}{x^{2}}\right ) x^{2}}{\left (x^{2}+1\right ) \ln \left (\frac {1}{x}\right )}+\frac {\ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}+\frac {\ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )^{2}}\\ f_1 f_2 &=\frac {\left (-\ln \left (\frac {x^{2}+1}{x}\right ) x -\ln \left (\frac {1}{x}\right )\right ) \ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )^{2}}\\ f_2^2 f_0 &=0 \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x u^{\prime \prime }\left (x \right )}{\ln \left (\frac {1}{x}\right )}-\left (\frac {\left (2-\frac {x^{2}+1}{x^{2}}\right ) x^{2}}{\left (x^{2}+1\right ) \ln \left (\frac {1}{x}\right )}+\frac {\ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}+\frac {\ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )^{2}}+\frac {\left (-\ln \left (\frac {x^{2}+1}{x}\right ) x -\ln \left (\frac {1}{x}\right )\right ) \ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )^{2}}\right ) u^{\prime }\left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} +\left (\int {\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right )^{2} x^{3}+\ln \left (\frac {x^{2}+1}{x}\right )^{2} x -\ln \left (\frac {x^{2}+1}{x}\right ) x^{2}-x^{2} \ln \left (\frac {1}{x}\right )-\ln \left (\frac {x^{2}+1}{x}\right )+\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right ) \left (x^{2}+1\right ) \ln \left (\frac {x^{2}+1}{x}\right )}d x \right )}d x \right ) c_{2} \] The above shows that \[ u^{\prime }\left (x \right ) = {\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right )^{2} x^{3}+\ln \left (\frac {x^{2}+1}{x}\right )^{2} x -\ln \left (\frac {x^{2}+1}{x}\right ) x^{2}-x^{2} \ln \left (\frac {1}{x}\right )-\ln \left (\frac {x^{2}+1}{x}\right )+\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right ) \left (x^{2}+1\right ) \ln \left (\frac {x^{2}+1}{x}\right )}d x \right )} c_{2} \] Using the above in (1) gives the solution \[ y = -\frac {{\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right )^{2} x^{3}+\ln \left (\frac {x^{2}+1}{x}\right )^{2} x -\ln \left (\frac {x^{2}+1}{x}\right ) x^{2}-x^{2} \ln \left (\frac {1}{x}\right )-\ln \left (\frac {x^{2}+1}{x}\right )+\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right ) \left (x^{2}+1\right ) \ln \left (\frac {x^{2}+1}{x}\right )}d x \right )} c_{2} \ln \left (\frac {1}{x}\right )}{\ln \left (\frac {x^{2}+1}{x}\right ) x \left (c_{1} +\left (\int {\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right )^{2} x^{3}+\ln \left (\frac {x^{2}+1}{x}\right )^{2} x -\ln \left (\frac {x^{2}+1}{x}\right ) x^{2}-x^{2} \ln \left (\frac {1}{x}\right )-\ln \left (\frac {x^{2}+1}{x}\right )+\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right ) \left (x^{2}+1\right ) \ln \left (\frac {x^{2}+1}{x}\right )}d x \right )}d x \right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {{\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right )^{2} x^{3}+\ln \left (\frac {x^{2}+1}{x}\right )^{2} x -\ln \left (\frac {x^{2}+1}{x}\right ) x^{2}-x^{2} \ln \left (\frac {1}{x}\right )-\ln \left (\frac {x^{2}+1}{x}\right )+\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right ) \left (x^{2}+1\right ) \ln \left (\frac {x^{2}+1}{x}\right )}d x \right )} \ln \left (\frac {1}{x}\right )}{\ln \left (\frac {x^{2}+1}{x}\right ) x \left (c_{3} +\int {\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right )^{2} x^{3}+\ln \left (\frac {x^{2}+1}{x}\right )^{2} x -\ln \left (\frac {x^{2}+1}{x}\right ) x^{2}-x^{2} \ln \left (\frac {1}{x}\right )-\ln \left (\frac {x^{2}+1}{x}\right )+\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right ) \left (x^{2}+1\right ) \ln \left (\frac {x^{2}+1}{x}\right )}d x \right )}d x \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {{\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right )^{2} x^{3}+\ln \left (\frac {x^{2}+1}{x}\right )^{2} x -\ln \left (\frac {x^{2}+1}{x}\right ) x^{2}-x^{2} \ln \left (\frac {1}{x}\right )-\ln \left (\frac {x^{2}+1}{x}\right )+\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right ) \left (x^{2}+1\right ) \ln \left (\frac {x^{2}+1}{x}\right )}d x \right )} \ln \left (\frac {1}{x}\right )}{\ln \left (\frac {x^{2}+1}{x}\right ) x \left (c_{3} +\int {\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right )^{2} x^{3}+\ln \left (\frac {x^{2}+1}{x}\right )^{2} x -\ln \left (\frac {x^{2}+1}{x}\right ) x^{2}-x^{2} \ln \left (\frac {1}{x}\right )-\ln \left (\frac {x^{2}+1}{x}\right )+\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right ) \left (x^{2}+1\right ) \ln \left (\frac {x^{2}+1}{x}\right )}d x \right )}d x \right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {{\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right )^{2} x^{3}+\ln \left (\frac {x^{2}+1}{x}\right )^{2} x -\ln \left (\frac {x^{2}+1}{x}\right ) x^{2}-x^{2} \ln \left (\frac {1}{x}\right )-\ln \left (\frac {x^{2}+1}{x}\right )+\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right ) \left (x^{2}+1\right ) \ln \left (\frac {x^{2}+1}{x}\right )}d x \right )} \ln \left (\frac {1}{x}\right )}{\ln \left (\frac {x^{2}+1}{x}\right ) x \left (c_{3} +\int {\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right )^{2} x^{3}+\ln \left (\frac {x^{2}+1}{x}\right )^{2} x -\ln \left (\frac {x^{2}+1}{x}\right ) x^{2}-x^{2} \ln \left (\frac {1}{x}\right )-\ln \left (\frac {x^{2}+1}{x}\right )+\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right ) \left (x^{2}+1\right ) \ln \left (\frac {x^{2}+1}{x}\right )}d x \right )}d x \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{2} \ln \left (\frac {x^{2}+1}{x}\right ) x^{2}-y \ln \left (\frac {x^{2}+1}{x}\right ) x -y^{\prime } x \ln \left (\frac {1}{x}\right )-y \ln \left (\frac {1}{x}\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {-y^{2} \ln \left (\frac {x^{2}+1}{x}\right ) x^{2}+y \ln \left (\frac {x^{2}+1}{x}\right ) x +y \ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 95
dsolve(diff(y(x),x) = y(x)*(-ln(1/x)-ln((x^2+1)/x)*x+ln((x^2+1)/x)*x^2*y(x))/x/ln(1/x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x \right )}}{-\left (\int \frac {{\mathrm e}^{-\left (\int \frac {\ln \left (\frac {x^{2}+1}{x}\right ) x +\ln \left (\frac {1}{x}\right )}{x \ln \left (\frac {1}{x}\right )}d x \right )} x \ln \left (\frac {x^{2}+1}{x}\right )}{\ln \left (\frac {1}{x}\right )}d x \right )+c_{1}} \]
✓ Solution by Mathematica
Time used: 0.91 (sec). Leaf size: 110
DSolve[y'[x] == (y[x]*(-Log[x^(-1)] - x*Log[(1 + x^2)/x] + x^2*Log[(1 + x^2)/x]*y[x]))/(x*Log[x^(-1)]),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\exp \left (\int _1^x\left (-\frac {\log \left (K[1]+\frac {1}{K[1]}\right )}{\log \left (\frac {1}{K[1]}\right )}-\frac {1}{K[1]}\right )dK[1]\right )}{-\int _1^x\frac {\exp \left (\int _1^{K[2]}\left (-\frac {\log \left (K[1]+\frac {1}{K[1]}\right )}{\log \left (\frac {1}{K[1]}\right )}-\frac {1}{K[1]}\right )dK[1]\right ) K[2] \log \left (K[2]+\frac {1}{K[2]}\right )}{\log \left (\frac {1}{K[2]}\right )}dK[2]+c_1} \\ y(x)\to 0 \\ \end{align*}