Internal problem ID [9023]
Internal file name [OUTPUT/7958_Monday_June_06_2022_01_01_13_AM_85806484/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 689.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati", "homogeneousTypeD2"
Maple gives the following as the ode type
[[_homogeneous, `class D`], _Riccati]
\[ \boxed {y^{\prime }-\frac {x y-y-{\mathrm e}^{x +1} x^{3}+{\mathrm e}^{x +1} x y^{2}}{\left (x -1\right ) x}=0} \]
Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} {\mathrm e}^{x +1} x^{3} u \left (x \right )^{2}-{\mathrm e}^{x +1} x^{3}-\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x^{2}+x^{2} u \left (x \right )+\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x -u \left (x \right ) x = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {{\mathrm e}^{x +1} x \left (u^{2}-1\right )}{x -1} \end {align*}
Where \(f(x)=\frac {{\mathrm e}^{x +1} x}{x -1}\) and \(g(u)=u^{2}-1\). Integrating both sides gives \begin{align*} \frac {1}{u^{2}-1} \,du &= \frac {{\mathrm e}^{x +1} x}{x -1} \,d x \\ \int { \frac {1}{u^{2}-1} \,du} &= \int {\frac {{\mathrm e}^{x +1} x}{x -1} \,d x} \\ -\operatorname {arctanh}\left (u \right )&={\mathrm e}^{x +1}-{\mathrm e}^{2} \operatorname {expIntegral}_{1}\left (1-x \right )+c_{2} \\ \end{align*} The solution is \[ -\operatorname {arctanh}\left (u \left (x \right )\right )-{\mathrm e}^{x +1}+{\mathrm e}^{2} \operatorname {expIntegral}_{1}\left (1-x \right )-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} -\operatorname {arctanh}\left (\frac {y}{x}\right )-{\mathrm e}^{x +1}+{\mathrm e}^{2} \operatorname {expIntegral}_{1}\left (1-x \right )-c_{2} = 0\\ -\operatorname {arctanh}\left (\frac {y}{x}\right )-{\mathrm e}^{x +1}+{\mathrm e}^{2} \operatorname {expIntegral}_{1}\left (1-x \right )-c_{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\operatorname {arctanh}\left (\frac {y}{x}\right )-{\mathrm e}^{x +1}+{\mathrm e}^{2} \operatorname {expIntegral}_{1}\left (1-x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ -\operatorname {arctanh}\left (\frac {y}{x}\right )-{\mathrm e}^{x +1}+{\mathrm e}^{2} \operatorname {expIntegral}_{1}\left (1-x \right )-c_{2} = 0 \] Verified OK.
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x y -y -{\mathrm e}^{x +1} x^{3}+{\mathrm e}^{x +1} x \,y^{2}}{\left (x -1\right ) x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {y}{x -1}-\frac {y}{\left (x -1\right ) x}-\frac {x^{2} {\mathrm e}^{x} {\mathrm e}}{x -1}+\frac {{\mathrm e}^{x} {\mathrm e} y^{2}}{x -1} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {x^{2} {\mathrm e}^{x +1}}{x -1}\), \(f_1(x)=\frac {1}{x}\) and \(f_2(x)=\frac {{\mathrm e}^{x +1}}{x -1}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {{\mathrm e}^{x +1} u}{x -1}} \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 {{\mathrm e}^{x +1}}{x -1}-\frac {{\mathrm e}^{x +1}}{\left (x -1\right )^{2}}\\ f_1 f_2 &=\frac {{\mathrm e}^{x +1}}{x \left (x -1\right )}\\ f_2^2 f_0 &=-\frac {{\mathrm e}^{3 x +3} x^{2}}{\left (x -1\right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {{\mathrm e}^{x +1} u^{\prime \prime }\left (x \right )}{x -1}-\left (\frac {{\mathrm e}^{x +1}}{x -1}-\frac {{\mathrm e}^{x +1}}{\left (x -1\right )^{2}}+\frac {{\mathrm e}^{x +1}}{x \left (x -1\right )}\right ) u^{\prime }\left (x \right )-\frac {{\mathrm e}^{3 x +3} x^{2} u \left (x \right )}{\left (x -1\right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{i \left (\int \frac {x \,{\mathrm e}^{x +1} \tan \left (i \operatorname {expIntegral}_{1}\left (1-x \right ) {\mathrm e}^{2}-i {\mathrm e}^{x +1}+c_{1} \right )}{x -1}d x \right )} c_{2} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {i c_{2} \tan \left (i \operatorname {expIntegral}_{1}\left (1-x \right ) {\mathrm e}^{2}-i {\mathrm e}^{x +1}+c_{1} \right ) x \,{\mathrm e}^{x +1+i \left (\int \frac {x \,{\mathrm e}^{x +1} \tan \left (i \operatorname {expIntegral}_{1}\left (1-x \right ) {\mathrm e}^{2}-i {\mathrm e}^{x +1}+c_{1} \right )}{x -1}d x \right )}}{x -1} \] Using the above in (1) gives the solution \[ y = -i \tan \left (i \operatorname {expIntegral}_{1}\left (1-x \right ) {\mathrm e}^{2}-i {\mathrm e}^{x +1}+c_{1} \right ) x \,{\mathrm e}^{x +1+i \left (\int \frac {x \,{\mathrm e}^{x +1} \tan \left (i \operatorname {expIntegral}_{1}\left (1-x \right ) {\mathrm e}^{2}-i {\mathrm e}^{x +1}+c_{1} \right )}{x -1}d x \right )} {\mathrm e}^{-1-x} {\mathrm e}^{\int -\frac {i x \,{\mathrm e}^{x +1} \tan \left (i \operatorname {expIntegral}_{1}\left (1-x \right ) {\mathrm e}^{2}-i {\mathrm e}^{x +1}+c_{1} \right )}{x -1}d x} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -i x \tan \left (i \operatorname {expIntegral}_{1}\left (1-x \right ) {\mathrm e}^{2}-i {\mathrm e}^{x +1}+c_{3} \right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -i x \tan \left (i \operatorname {expIntegral}_{1}\left (1-x \right ) {\mathrm e}^{2}-i {\mathrm e}^{x +1}+c_{3} \right ) \\ \end{align*}
Verification of solutions
\[ y = -i x \tan \left (i \operatorname {expIntegral}_{1}\left (1-x \right ) {\mathrm e}^{2}-i {\mathrm e}^{x +1}+c_{3} \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {\mathrm e}^{x +1} x y^{2}-{\mathrm e}^{x +1} x^{3}-y^{\prime } x^{2}+x y+y^{\prime } x -y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{x +1} x^{3}-{\mathrm e}^{x +1} x y^{2}-x y+y}{-x^{2}+x} \end {array} \]
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 homogeneous D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 25
dsolve(diff(y(x),x) = (x*y(x)-y(x)-exp(x+1)*x^3+exp(x+1)*x*y(x)^2)/(x-1)/x,y(x), singsol=all)
\[ y \left (x \right ) = -\tanh \left ({\mathrm e}^{x +1}-{\mathrm e}^{2} \operatorname {expIntegral}_{1}\left (1-x \right )+c_{1} \right ) x \]
✓ Solution by Mathematica
Time used: 0.853 (sec). Leaf size: 67
DSolve[y'[x] == (-(E^(1 + x)*x^3) - y[x] + x*y[x] + E^(1 + x)*x*y[x]^2)/((-1 + x)*x),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x-x e^{2 \left (e^2 \operatorname {ExpIntegralEi}(x-1)+e^{x+1}+c_1\right )}}{1+e^{2 \left (e^2 \operatorname {ExpIntegralEi}(x-1)+e^{x+1}+c_1\right )}} \\ y(x)\to -x \\ y(x)\to x \\ \end{align*}